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University  of  California, 


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A   TREATISE 


ON 


PLANE  AND   SPHERICAL 

TRIGONOMETRY; 


NCLUDING    THE 


CONSTRUCTION  OF  THE  AUXILIARY  TABLES; 


A     CONCISE 


TRACT  ON  THE  CONIC  SECTIONS, 


AND     THE 


PRINCIPLES  OF  SPHERICAL  PROJECTION. 


BY   ENOCH   LEWIS. 


PHILADELPHIA: 
URIAH    HUNT    &    SON, 

No.  62  NORTH  FOUXITII  STltEUT. 

1860. 


r 


Entered,  according  to  the  Act  ot  Congress,  in  the  year  1844,  by 

HECTOR    ORR, 

in  the  clerk's  office  of  the  District  Court  of  the  United  States  in  and  for  the 
Eastern  District  of  Pennsylvania. 


J.  Pagan,  Stereotyper. 


PREFACE. 

It  will  probably  appear,  to  some,  a  work  of  super- 
erogation to  add  another  tract  on  Plane  and  Spherical 
Trigonometry  to  the  great  number  already  before  the 
public ;  especially  as  some  of  those  treatises  are  the 
productions  of  men  whose  talents  and  attainments 
were  unquestionably  of  the  highest  order.  Still,  it  has 
appeared  to  me  that,  however  valuable  many  of  those 
works  must  be  considered,  there  are  none  of  them 
exactly  suited  to  the  use  of  schools  in  which  this 
branch  of  mathematics  is  traced  to  its  principles.  It 
is,  indeed,  no  unusual  thing  to  find  young  men  who 
have  studied  this  science  in  the  way  it  is  commonly 
taught,  who  are  very  imperfectly  acquainted  with  the 
nature,  and  almost  entirely  ignorant  of  the  construc- 
tion, of  the  tables  which  they  are  continually  using. 
And  it  must  be  admitted,  that,  when  the  nature  and 
construction  of  logarithms,  and  of  sines  and  tangents, 
are  explained  by  Algebra  and  common  Geometry,  the 
processes  are  generally  either  so  obscure,  or  so  prolix, 
as  to  discourage  the  majority  of  students.  The  Diffe- 
rential Calculus  is  well  known  to  furnish  the  most 
direct,  if  not  the  only  direct,  and  simple  method  of 

A*  (v) 


vi  PREFACE. 

investigating  the  formulae  by  which  those  tables  are 
most  expeditiously  computed.  But  that  calculus  itself, 
as  commonly  exhibited,  presents  so  many  refined 
speculations,  that  very  few,  except  those  who  have  a 
taste  for  mathematical  studies,  can  avail  themselves  of 
its  advantages. 

As  it  is  evidently  unscientific  to  erect  a  system, 
either  in  theory  or  practice,  upon  unknown  principles, 
it  has  been  my  object,  in  the  following  work,  to  trace 
every  process  which  is  required  to  be  adopted,  to 
principles  which  are  supposed  to  be  previously  under- 
stood. The  student  is  supposed  to  be  already 
acquainted  with  Algebra  and  Geometry.  If  the 
student  is  master  of  the  first  six,  and  the  eleventh 
books  of  Euclid ;  or,  which  is  nearly  the  same  thing, 
of  the  first  six,  and  the  second  supplementary  book 
of  Playfair's  Geometry ;  and  of  as  much  algebra  as  is 
contained  in  the  first  ninety  pages  of  my  treatise ;  he 
may  proceed  with  confidence  to  the  study  of  the 
following  tract. 

This  work  was  intended  to  include  as  much  only 
of  the  Differential  Calculus,  as  the  elucidation  of  the 
science  of  Trigonometry  required.  I  have  therefore 
confined  myself  to  differentials  of  the  first  order;  and, 
by  the  use  of  proper  expedients,  have  deduced  the 
requisite  formulae  from  those  differentials.     Some  of 


PREFACE.  vii 

the  methods  used  in  this  work  are  supposed  to  be  new  ; 
and,  if  so,  they  may  be  considered  as  improvements 
upon  the  labours  of  my  predecessors.  Of  this  cha- 
racter are  the  investigations  of  Gregory's  theorems 
for  computing  an  arc  in  terms  of  its  tangent,  and  for 
computing  logarithms. 

The  treatises  on  Spherical  Trigonometry  with  which 
our  schools  are  supplied,  are  nearly  all  of  them  desti- 
tute of  anything  on  the  subject  of  Spherical  Projec- 
tions. This  appears  to  me  an  important  defect.  A 
small  tract  on  that  subject,  which  I  added,  more  than 
thirty  years  ago,  to  a  Philadelphia  edition  of  Thomas 
Simpson's  Plane  and  Spherical  Trigonometry,  is  the 
only  one,  so  far  as  I  know,  which  is  to  be  found  in 
our  schools ;  unless  we  consider  Davies's  Descriptive 
Geometry  as  one.  Simpson's  work  being  now  out  of 
print,  and  the  work  of  Davies,  notwithstanding  its 
merits,  not  appearing  calculated  to  supply  the  place  of 
the  appendix,  I  have  revised,  or  rather  written  anew, 
that  part  of  my  early  labours,  and  subjoined  it  to  this 
work. 

The  present  treatise  being  designed  as  an  introduc- 
tion or  preliminary  to  Astronomy,  a  concise  tract  on 
the  Conic  Sections,  including  all  the  properties  of  the 
ellipse  and  parabola  which  are  usually  cited  by  writers 
on  that  science,  has  been  introduced.     This  appeared 


viii  PREFACE. 

requisite,  because  some  of  those  properties  were 
unavoidably  referred  to  in  the  tract  on  Spherical 
Projections ;  and,  among  the  treatises  on  that  subject 
already  in  print,  it  was  not  easy  to  fix  upon  any  one 
to  which  I  could  refer,  that  would  be  generally  known 
to  my  readers.  Besides,  it  appeared  no  difficult  matter 
to  include  in  this  work  all  the  information  which  the 
astronomic  inquirer  would  need  in  the  prosecution  of 
his  studies.  The  Conic  Sections  being,  as  the  name 
implies,  derived  from  the  cutting  of  a  cone,  I  thought 
it  more  direct  to  deduce  the  primary  propositions  from 
the  section  of  the  cone,  than  to  lay  down  first  a  plane 
figure,  derived  in  a  different  way;  and,  after  demon- 
strating most  of  its  properties,  to  prove  at  last  its 
identity  with  a  conic  section. 

In  the  practical  examples,  some  astronomical  terms 
are  used  which  are  not  defined,  because  it  was  sup- 
posed that  few  young  persons  would  study  this  work 
without  some  previous  acquaintance  with  terms  so 
generally  understood. 

The  references  to  the  properties  of  geometrical 
figures  are  made  to  Playfair's  Euclid ;  but  they  will 
generally  apply  to  Simson's  translation  of  Euclid's 
Elements.  This  oldest  work  on  Geometry,  with  the 
few  improvements  introduced  by  the  Scotch  geometer, 
is,  in  my  opinion,  better  calculated  to  lead  the  attentive 


PREFACE.  ix 

student  into  an  accurate  acquaintance  with  this  noble 
science,  than  any  modern  treatise  which  has  fallen  into 
my  way.  Among  those  improvements,  however,  the 
substitution  of  the  language  of  Algebra,  in  the  fifth 
book,  in  place  of  that  which  Euclid  made  use  of,  Can 
hardly  be  reckoned  as  one.  Play  fair's  exposition  of 
the  fifth  definition  is  certainly  a  good  one;  but,  in 
other  respects,  I  consider  Simson's  translation  of  that 
book  greatly  preferable  to  the  form  in  which  Playfair 
has  left  it. 

Philadelphia,  10  Mo.  1844. 


PLANE  TRIGONOMETRY. 


INTRODUCTION. 

In  the  practice  of  Trigonometry  there  are  several  tables 
generally  used,  the  construction  and  uses  of  which  constitute 
an  essential  part  of  the  science.  But  when  the  construction 
of  these  tables  is  deduced  from  Geometry  and  common  Alge- 
bra, the  subject  is  certainly  presented  to  the  student  in  a  very 
discouraging  shape.  The  rules  by  which  these  tables  are 
most  readily  computed,  are  easily  derived  from  the  Differen- 
tial Calculus ;  but  that  branch  of  science,  when  pursued  to 
any  considerable  extent,  involves  many  refined  and  difficult 
speculations.  It  therefore  generally  happens  that  the  stu- 
dents of  Trigonometry  acquire  a  practical  acquaintance  with 
the  auxiliary  tables,  but  understand  neither  their  construc- 
tion nor  nature. 

As  those  parts  of  the  Differential  Calculus  which  the  con- 
struction of  all  the  tables  commonly  used  in  trigonometrical 
calculations  absolutely  demands,  lie  within  a  narrow  compass, 
and  involve  no  very  difficult  inquiries,  I  shall  employ  a  few 
pages  in  explaining  the  elements  of  this  science,  so  as  to 
enable  the  student  to  understand  the  nature  and  origin  of  the 
formulae  which  are  commonly  used  in  the  computation  of  the 
trigonometrical  tables 

CM) 


12  PLANE  TRIGONOMETRY. 

Article  1.  The  Differential  Calculus  is  founded  essentially 
upon  the  relation  which  variable  and  dependent  quantities, 
considered  as  decreasing  till  they  vanish,  bear  to  each  other 
in  their  evanescent  or  vanishing  state.  This  rati-o  is  called 
their  ultimate  ratio.  Newton  observes  that  the  ultimate  ratio 
of  evanescent  quantities  is  not  the  ratio  which  they  have 
before  they  vanish,  nor  afterwards  ;  but  the  ratio  with  which 
they  vanish.*  As  the  ultimate  ratio  of  vanishing  quantities 
is  not  the  ratio  which  they  have  before  they  vanish,  but  at 
the  instant  when  they  vanish,  it  is  most  convenient  to  deter- 
mine that  ratio  by  supposing  the  evanescent  quantities  to 
have  actually  vanished.  This  may  often  be  done  in  a  way 
which  leaves  no  room  for  error  or  doubt.  One  or  two  exam- 
ples in  common  Algebra  will  render  this  obvious : 


xr 

— —a+x 


fi  —  x? 

— — =  aiJraxJrx~. 


Now,  these  results  being  strictly  correct,  whatever  value 
may  be  assigned  to  a  or  x<  let  x  be  supposed  at  first  less  than 
a,  and  to  increase  till  it  becomes  equal  to  a ;  d1  —  x2  and 
a  —  x,  are  thus  rendered  decreasing  and  evanescent  quanti- 
ties, whose  ultimate  ratio  is  the  ratio  which  they  have,  not 
before  x  becomes  equal  to  a,  but  at  the  instant  when  that 

equality  is   attained.      But  when  x=a,  a+x=2a=-y  ;  and 

ar-\-ax-\-x2=3a2=-^-.     Hence  the  ultimate  ratio  of  a  —  x  to 

a-  —  or,  when  x  approximates  and  ultimately  equals  a,  is  the 
ratio  of  1  :  2a;  and  the  ultimate  ratio  of  a  —  x  to  aa  —  x3, 
under  like  circumstances,  is  the  ratio  of  1  :  3a\ 


Principia,  Scholium,  Book  I.,  Art.  1. 


INTRODUCTION.  13 

Art.  2.  In  the  investigations  connected  with  the  Differen- 
tial Calculus,  quantities  are  considered  under  two  very  dif- 
rent  aspects:  constant  and  variable. 

Constant  quantities  are  such  as  remain  unchanged,  or 
retain  the  same  value  throughout  the  investigation.  Variable 
quantities  are  those  which  change  their  value  in  the  course 
of  the  solution  or  demonstration.  Constant  quantities  are 
usually  denoted  by  the  initial  letters,  a,  b,  c,  &c. ;  variable 
ones  by  the  final  letters,  x,  y,  z,  &c. 

Art.  3.  When  two  variable  quantities  enter  into  an  equa- 
tion, so  that  the  value  of  one  depends  upon  the  value  of  the 
other,  and  one  of  them  is  increased  by  any  quantity  or  incre- 
ment ;  the  quantity  which  must  be  added  to  the  other  to 
preserve  the  equation,  is  called  the  corresponding  increment. 
Thus,  if  y=ax2,  and  we  increase  x  by  the  increment  h,  so  that 
x  becomes  x-\-h,  the  new  value  of  y  will  be 
a(x+hy=ax2  +  2axh  +  ah2', 
hence  the  corresponding  increments  of  x  and  y  are  //  and 
2axh  -f  all2. 

The  variable  quantity  from  which  a  differential  is  deduced 
is  called  an  integral. 

Differentials  of  dependent  variables  may  be  considered  as 
the  corresponding  increments  of  those  variables,  in  the  ulti- 
mate state  of  the  increments,  when  they  are  diminished  till 
they  vanish.  Or  differentials  may  be  defined  to  be  quantities 
having  to  each  other  the  ratio  which  is  the  ultimate  ratio 
of  the  corresponding  increments,  those  increments  being  sup- 
posed to  decrease  till  they  vanish. 

Hence  it  follows  that  one  differential  cannot  enter  into 
an  equation  without  another.  We  never  inquire  into  the 
absolute,  but  merely  into  the  relative  values  of  differentials. 

Art.  4.  The  differential  of  a  variable  quantity  is  denoted 
by  prefixing  the  letter  d ;  thus  dx  signifies  the  differential  of 
x.  When  the  integral  is  a  compound  quantity,  it  is  enclosed 
in  brackets,  or  a  point  is  introduced  between  it  and  the  letter 
d :  thus  the  differential  of  xn  is  expressed  by  d(xn),  or  d.xn ; 

B 


14  PLANE  TRIGONOMETRY. 

whereas  dxn  signifies  the  n  power  of  dx,  the  same  as  {dx)n. 
The  differential  of  x+y  is  also  expressed  by  d(x+y).  In  these 
cases  the  letter  d  is  not  an  algebraic  quantity,  but  a  sign,  or 
an  abridgement  of  the  words  differential  of. 

Art.  5.  Let  y  =  x  +  z,  and  let  x  and  z  be  increased  by  the 
increments  h  and  k  respectively.  Denote  the  new  value  of  y 
by  y' ;  so  that 

y  =  x  +  h  -\-  z  +  h ; 

wherefore  y  —  y  =  h  +  k; 

or  the  increment  of  y  =  the  sum  of  the  increments  of  x  and  z. 
Now,  this  being  true,  whether  the  increments  are  taken  in 
their  vanishing  state  or  any  other,  we  evidently  have 

dy  =  dx  +  dz. 

Again :  Let  y  =  ax, 

and  y'  =  a(x-\-h)  =  ax  +  ah ; 

then  y'  —  y  —  ah; 

which  is  also  true,  whether  the  increment  h  is  taken  in  its 
vanishing  or  finite  state :  consequently, 

dy  =  adx. 
Art.  6.  Let  y  —  x2, 

and  y'  =  (x  +  h)"  =  *'  +  %xh  +  &; 

whence  y'  —  y  =  2a:A  +  A2, 

and  y   ,    *=%x  -{■  h  = — , — : 

n  1 

consequently,         % ;  :  y  —  y  :  :  1  :  2#  +  A. 

But  the  ultimate  ratio  of  1  :  2x  +  h  is  the  ratio  of  1  :  2x ; 

wherefore,  dx  :  dy  :  :  1  :  2a: ; 

or     .         .         .         .        dy  =  2o%£r ; 

that  is,  da:2  =  2^^r.  If  the  equation  dy  =  2xdx  be  put  into 
this  form, 


INTRODUCTION.  15 

dx  ~  ** 
the  last  member  2x  is  called  the  differential  coefficient. 
Art.  7.  To  find  the  differential  of  xz. 
Put  y  = :  x  +  z ; 

then  y2  =  (a?  +  z)2  =  af  +  2xz  +  z2 ; 

and  (Art.  5,)  dy  =  dx  +  dz; 

also  c?.y2  =  d.(x  +  z)2  = 

(Art.  6)  2(x  +  z)  (c?x  +  dz)  =  2xdx  +  2xdz  +  2zdx  +  2z<7z. 
But  '     d.y1  =  d(x>  +  2xz  +  z2)  =* 

(Arts.  5  &  6)  2.^  +  2^(xz)  +  2zdz. 

Comparing  these  values  of  d.y2,  we  have 

2d.(xz)  =  2xdz  +  2zdx ; 
or  d.xz  —  xdz  +  zdx. 

Hence,         d.xyz  =  xd.yz  +  yzdx  ==  xz/^z  +  rrzdy  +  yzcfo. 

Again : 

d.xyz  _  dz      dy      dx 
xyz  z        y        x 

and  the  same  thing  may  be  proved,  whatever  be  the  number 
of  variables. 

Art.  8.  To  find  the  differential  of  xn,  n  being  a  whole 
positive  number. 

It  is  obvious  that  xn  =  x.x.x.x  to  n  terms.   Hence,  by  Art.  7, 

d.xn      dx      dx                                       ndx 
— —  = h  —  +  &c.  to  n  terms  =  : 

wherefore,  by  clearing  of  fractions, 

d.xn  p  nxxl~]dx. 

x 
Art.  9.  To  find  the  differential  of  —  ,  both  numerator  and 


denominator  being  variable. 


16  PLANE  TRIGONOMETRY. 

Put  z  =  -  ; 

y 

wlience,  y%  —  x, 

and,  (Art.  7),  ydz  +  zdy  =  dx : 

consequently, 

dx—^ 

d%  ^  ^—Z^M  = &  =  ydx  —  xdy 

y  y  yl  ' ,  J 


that  is,  d.'~  — 


ydx  —  xdy 


y  y 

m 

Art.  10.  To  find  d.x",  m  and  n  being  positive  integers. 

m 

Put  y  =  a? ; 

then,  .         .         .         yn  —  xm; 

and,  (Art.  8)  nyn~hly  =  mxm-}dx; 

whence, 

mxm~}dx       m  xm~l  m    —~\ 

Art.  11.  To  find  the  differential  of  a?_n  =  — . 

xn 

then  yxn  =  1. 

But  1  being  invariable,  its  differential  is  0;  therefore, 

d.yxn  =  0. 

Or,  yd.xn  -f  xndy  =  0  ; 

that  is,  nyxn~~]dx  +  xndy  =  0 : 

whence, 

j        — nyxn~xdx  xn~]dx      — ndx 

dy  = "Ji =  —  n-^ir  =^+r=—  nx  —  ^dx; 

or,  <7.ar~n  =  —  nx~'n~~]dx. 

From  this  article,  and  Arts.  8  and  10,  it  is  evident  that 


INTRODUCTION.  17 

d.xn  =  nxn'  xdx, 
whether  n  is  integral  or  fractional,  positive  or  negative. 

Art.  12.  Let  the  two  ascending  series,  Ax*  +  Bxb  +  Cxc  + 
Dxd  +  &c,  and  Mxm  +  Nxn  +  Pxp  +  Qx*  +  &c,  be  always 
equal ;  so  that  whatever  value  may  be  assigned  to  x,  we  shall 
still  have, 

Axa+Bx,b  +  Grs  +  Da:d+  &c.  =  Mxm  +  Na:n  +  Pa:p  +  Q^  +  &c. ; 
then,  a  =  m,  b  =  n,  c  =  p,  &c. ; 

and  A  -  M,  B  =  N,  C  =  P,  &c. 

For,  if  possible,  let  a  be  less  than  m,  and  divide  the  equation 
by  #a;  then 

A  +  Bo*"11  +  Gcc"a  +  Da0"8  +  &c.  = 
Mxm-&  +  Nx"-*  +  Pxp-a  +  Qx-q-a  +  &c. 
But  as  the  series  are  both  ascending  ones,  b,  c,  d>  m,  n,  p,  q, 
&c.  are  all  greater  than  a.  Hence,  if  x  =  0,  all  the  terms  of 
this  equation,  except  the  first,  will  vanish ;  hence  in  that  case 
A  =  0,  which  is  evidently  absurd.  Therefore,  a  is  not  less 
than  m  ;  and  in  a  similar  way  it  may  be  proved  that  m  is  not 
less  than  a;,  therefore  a  =  m,  and  the  above  equation  be- 
comes 

A  +  Bxb~*  +  Oc-a  +  Dxda  +  &c.  = 

M  +  Ntfn-m  +  P^p-m  +  Qx*~m  +  &c. 
And  making  x  =  0,  A  =  M.     Consequently,  A#a  =  Mxm ; 
wherefore, 

Bxh  +  Cx*  +  DxA  +  &c.  =  Nxn  +  Pa:p  +  Q^  +  &c. 
Then,  by  the  same  process  of  reasoning,  we  find  b  =  n;  and 
B  =  N.     Hence  the  proposition  is  manifest. 

Art.  13.  An  important  application  of  the  property  just 
announced  may  be  exhibited  in  the  demonstration  of  New- 
ton's binomial  theorem. 

It  is  evident  that  in  the  general  development  of  (1  -f-  x)n, 
the  first  term  must  be  1 ;  for  when  x  =  0,  (1  +  x)n  =  ln  =  1. 
We  may  therefore  assume 

(1  +  x)n  =  1  +  Axp  +  Bx«  +  CxT  +  Da?8  +  &c. 

3  B* 


18  PLANE   TRIGONOMETRY. 

in  which  A,  B,  &c.,  are  unknown,  but  determinate  coeffi- 
cients; and  p,  q,  r,  &c,  unknown  exponents,  integral  or 
fractional,  positive  or  negative.  Suppose  x  a  variable  quan- 
tity. Then,  differentiating  both  sides  of  this  equation,  and 
dividing  by  dx,  we  have,  % 

'    «.(1  -f  xf"1  =  pAx^  +  qBx-*-1  +  rCxT~l  +  sT>x*~l  -f  &c. 
Multiplying  bj  I  +  x, 

rc.(l  +  ;r)n  =  pAx»~l  +  qBx*"1  +  rCxr~l  +  sDx8  -1  +  &c. 
pAx»  +  qBx*  +  rCxT  +  s&x3  +  &c. 
Then,  from  first  equation,  multiplying  by  n, 

n{\  +  x)n  =  n  -f  nAxp  +  nBxq  +  nCxr  +  &c. 
These  equations  being  identical,  we  have  by  transposition, 

pAx'-]  +  qBx*-1  +  rCxT"1  +  sDx"1  +  &c.  =  n  + 
(n  —  p)  Axp  +  (n  —  q)  BxQ  +  (n  —  r)  CxT  +  (n  —  s)  ~Dx9  +  &c.  ; 
and  by  comparing,  first  the  exponents,  and  then  the  coeffi- 
cients, (Art.  12,)  we  have 

p  —  1  =  Q ;  q  —  1  =  p;  r  —  I  —  q ;  &c. ; 
or,  p  =  1 ;  q  =  2  ;  r  =  3  ;  s  =  4;  &c. : 

and 

a  t>     n—1  a        n  —  ln     n  —  %v>        n— In— 2 

A=w;B=— g—  A=n. — g-  ;C=  — ^—  B=n.—^-.—^-;  &c. 

Consequently, 

w — 1  c        w  —  In  —  2  , 
(1  +  x)n  =  1  +  n#  +  n. — Q—ar+  w-~o — • — q~~  ^    +  &c- 

From  this  we  readily  obtain  the  development  of  (a  +  by. 
For, 

(a+by  =  a«(l  +  -)"; 

in  which  we  have  —  instead  of  x  in  the  preceding.     Conse- 
quently, 
,       7V  (  ,  b  n  —  \  b*  n— In— 2  P  .      ) 


INTRODUCTION.  19 

=  an  +  nan-}b  +  n.  —^~an-2b2  +  n.  —^— .  — ^—  an~W  &c. 

As  the  equation  d.xn  —  nxn~~~^dx,  on  which  this  demonstration 
is  founded,  is  equally  correct  whether  n  is  integral  or  frac- 
tional, positive  or  negative,  it  is  evident  that  the  preceding 
development  of  (a  +  b)n  is  also  correct,  whatever  may  be  the 
value  of  n. 


Of  Logarithms. 

The  calculations  which  are  connected  with  Trigonometry 
are  much  facilitated  by  the  use  of  logarithms ;  it  will  there- 
fore be  proper,  in  a  treatise  on  that  science,  to  explain  their 
nature  and  use. 

Art.  14.  If  we  take  a  series  of  numbers  in  geometrical 
proportion,  beginning  with  a  unit,  as  1,  a,  a-,  a3,  a4,  a5,  afi,  a7, 
&c,  it  is  manifest  that  the  product  of  any  two  of  these  terms 
is  a  term  whose  exponent  is  the  sum  of  the  exponents  of  the 
factors.     Thus, 

a\a?  =a  a7 ;     am.an  =  am+\ 
If  then  A  =  am,  and  B  =  a", , 
AB  =  am+n : 
A2  =  A.A  =  am+m  =  a2ra : 


On  the  other  hand 


A3  =  am+m+m 

a=  a3m,  &c. 

id, 

a1       .  A 
^*;B 

am 
=  —  =  am-n; 

1            m 

A'2  =  a2; 

1             m 

A3  =  a3. 

Hence  it  appears  that  a,  being  assumed  equal  to  any  num- 
ber at  pleasure,  if  we  can  find  such  values  of  m,  n,  &c,  that 
am  =  A,  an  =  B,  &c,  A,  B,  being  given  numbers,  then  calling 
m  the  logarithm  of  A,  n  the  logarithm  of  B,  &c. ;  the  loga- 
rithm of  AB  will  be  the  sum  of  the  logarithms  of  A  and  B. 


20  PLANE  TRIGONOMETRY. 

The  logarithm  of  ~  will  be  the  logarithm  of  A  diminished  by 

the  logarithm  of  B.  In  other  words,  the  business  of  multiply- 
ing and  dividing  by  given  numbers  may  be  effected  by  the 
addition  and  subtraction  of  their  logarithms. 

As  am  =  A,  a  given  number ;  we  readily  perceive  that,  by 
assuming  different  values  of  a,  we  shall  change  the  value  of 
m ;  that  is,  we  shall  have  different  numbers  to  denote  the 
logarithm  of  a  given  number  A,  by  varying  the  value  of  a. 
Thus  it  appears  there  may  be  an  indefinite  variety  of  systems, 
according  to  the  various  values  which  may  be  taken  for  a. 
This  quantity  a  is  called  the  radix  or  base  of  the  system. 

Art.  15.  To  investigate  a  formula  by  which  the  logarithm 
of  any  given  number  may  be  computed,  we  may  assume 
ax  =  y ;  y  being  any  given  number  whatever ;  then  x  =  loga- 
rithm of  y  :  and  the  object  in  view  is  to  find  a  general  expres- 
sion for  x  in  terms  of  y.  If  we  suppose  x  to  be  variable,  it 
is  manifest  that  y  —  ax  must  also  be  variable. 

In  the  first  place,  if  x  —  0,  then  y  '=  a0  —  1,  whatever  value 
may  be  assigned  to  a ;  it  is  therefore  evident  that  the  loga- 
rithm of  1  is  0  in  every  system. 

Now,  let  y'  =  ax+h  =  ax.ah  =  yab : 

and,  to  reduce  this  second  member  to  a  more  manageable 
form,  put  1  +  b  =  a ; 

then,  (Art.  13,) 

^=y(l+5)fcf=y+y  |  hb+h.^bHh!1-^.  —  P  +  &c.  | 

Therefore, 

y-  -  y  =  y  \  hb  +  h~b-  +  h!^.  ~V  +  &c.  j  ; 

and,  consequently, 

y'  —  y         c.       h  — 1_„     h—   I  h  —  2, 


h 


(1       it  —  i7C,     n — i  it  —  ^.,      e       ) 


INTRODUCTION.  21 

Now,  when  h  =  0,  the  series  in  the  second  member  of  this 
equation  becomes 

b  —  W  +  W  —  W  +  l^5  — &c.; 
and  this  is  the  value  to  which  this  series  approximates  as  h 
is  diminished,  and  to  which  it  arrives  only  at  the  instant 
when  h  =  0.     Put,  then, 

b-W  +  ±V-\V  +  #J.  &c.  =  ~; 

and  it  will  be  V_Zll  =  1-. 

h  m 

Hence  the  ultimate  ratio  of  h  to  y  —~  y  is  the  ratio  of  m  to  y, 

dy     y 

consequently,  j-  =s  — ; 

or, =4i'i&  (A) 

In  this  equation,  the  value  of  m  depends  upon  the  value  of  a ; 
and  as  a  may  be  assumed  at  pleasure,  we  may  assign  any 
value  we  please  to  m.  This  is  more  convenient  than  to  as- 
sume a  value  of  a,  and  from  that  assumption  to  find  the  value 
of  m.  It  is  usual  to  call  m  the  modulus  of  the  system.  When 
m  is  taken  =  1,  the  logarithms  thence  deduced  are  called 
hyperbolic  logarithms,  because  they  correspond  with  certain 
areas  contained  between  the  curve  and  asymptotes  of  an  equi- 
lateral hyperbola.  In  Briggs',  or  the  common  logarithms, 
the  radix  a  is  assumed  =  10 ;  but  m  is  computed  by  a  method 
hereafter  explained. 

Art.  16.  As  no  general  method  has  been  discovered  by 
which  to  express  the  logarithm  of  a  number  in  finite  terms 
of  the  number  itself,  we  are  obliged  to  have  recourse  to  infi- 
nite series.  When  numbers  are  to  be  computed  by  means  of 
such  series,  it  is  of  importance  to  have  the  series  constructed 
in  such  manner  that  the  successive  terms  shall  become  smaller 
and  smaller ;  so  that,  a  limited  number  of  terms  being  intro- 


22  PLANE   TRIGONOMETRY. 

duced  into  the  computation,  the  rest  of  the  Series  may  be 
rejected  without  sensible  error. 

To  find  the  logarithm  of  a  +  *,. a  being  constant,  and  z 
variable. 

By  Art.  15,  if x  —  log.  of?/, 

mdxi 
dx=  — -. 

y 

Assume,  then,  log.  of  a  +  z  =  log.  a  +  Azn  +  Bzp  -f  Czq+Dzr+ 
&c. ;  in  which  the  exponents  to,  p,  q,  &c,  as  well  as  the  co- 
efficients A,  B,  C,  &c,  are  indeterminate.  In  this  equation, 
if  z  =  0,  we  have  log.  a  =  log.  a,  as  it  evidently  ought  to  be ; 
and  the  quantities  to,  p,  q,  A,  B,  C,  &c,  being  susceptible  of 
any  value,  positive  or  negative,  integral  or  fractional,  the 
above  equation  must  express  the  log.  of  a  +  z,  if  it  can  be 
expressed  at  all  in  terms  of  z. 

Differentiating  this  equation,  and  dividing  by  dz,  we  have 

7YI 

■ — —  =  nAzn~l  +  pBzP"1  +  qCz*-x  +  rDz'"1  +  &c. 

a+z  i       ■ 

By  multiplication  and  transposition, 

naAzn-~l  +  poBz"™1  +  qaCz«-~l  +  raDzH  +  &c.  )  _  Q 
—  m-\-  nkzn  +  pBzp  +  qCzq  +  &c.  I  ~ 

Equating  the  exponents  and  the  coefficients  respectively  of 
the  corresponding  terms, 

to  —  1  =  0,  jo  —  1  =  to,  q  —  I  =p,  r  —  I  =  q,  &c. 
togA  —  m=0,  paB+TOA=0,  qaC  +  pB=0,  rdD  +  ^C=0,  &c. 
Hence,  to  =  1,  p  =  2,  q  =  3,  &c. ; 

and  A  =  -,B=  —  -—0=^-^,0  = — -<— ;,  &c. 

a  2ar  Sa*  4a{ 

Consequently, 

*•:■■>  ,  TOZZ  TO2Z2  TO2Z3         7TOZ4 

log.  («  +  ,)  =  log.  a  +  --  2^  +  3^-3^+  &c. 
Putting  —  z  for  -f  z, 


INTRODUCTION.  23 

,  •                 x       ,                mz        mz1       mz']       mzi        £ 
log.  (a  —  z)  ==  log.  a -— ,  —  5-3 j-j  —  &c. 

Therefore,        log.  --  =  2m  i  —  +  ~  +  -r— ( +  &c. 

°   a  —  z  la        3(1*        5a'J 


When  a  =  1, 

iog.;_±Hmj2+!+!+f+&c. 

To  find  a  number  in  terms  of  its  hyperbolic  log. 
In  this  case,  if  x  =  log,  y, 

?  =  </• 

dx 
Put  y  =  1  +  Axn  +  Bxp  +  O1  -{-  D,xr  +  &c. ;  where  y  =  1, 
when  a?  =  0,  as  it  ought  to  be. 

Differentiate  and  divide  by  dx ;  and 

J?  (=  y)  =  nAxn~l  +  pBx*-1  +  qCx*~l  +  rDaH  +  &c. 

Comparing  these  values  of  y,  and  equating  the  coefficients  and 
exponents  respectively,  we  have 

Art.  17.  From  the  formulae  contained  in  the  last  article, 
the  logarithms  of  small  numbers?  are  readily  computed ;  and 
those  of  large  ones  are  easily  deduced  from  the  smaller.  A 
few  examples  are  subjoined. 

Required  the  logarithm  of  2. 

Here  *  =  ~j  =  J. 

z  =  §  =  .3333333333 
^W|s  =.0370370370 
2s  =  j^=.  0041 152263 

z7  =  i  zs  =  .0004572474 
z9  =     .0000508053 


24 


And 


►LANE 

TRIGONOMETRY. 

z"  = 

.0000056450 

z13= 

6272 

Zl5  = 

697 

Z17= 

77 

%   a 

3333333333 

^  = 

123456790 

1^5  — 

8230453 

4%7= 

653211 

iz9= 

56450 

1  1 

5132 

1  ~13_ 

483 

_1  -15  = 

1  5Z      — 

46 

1  ~17_ 

5 

.3465735903 
Therefore,  the  log.  of  2  =  .6931471806  m;  and  log.  of  8  = 
2.0794415418  m. 

Next,  let  the  log.  of  f  be  required.     Here, 

This  number  being  substituted  for  z  in  the  foregoing  equa- 
tion, produces 

log.  off  =  . 2231435514m. 
Then  the  sum  of  these  logarithms  is  the 

log.  of  8*x  I  =  10. 
Therefore,  the      log.  of  10  =  2.3025850932  m. 

If  now  we  desire  to  compute  the  common  logarithm  of  any 
number,  the  radix  of  that  system  being  10,  the  log.  of  10  is 
=  1 :  putting,  therefore,  the  log.  of  10  just  found  equal  1, 
we  find, 

m  =  OTS0S33  =  •4342944819" 

The  coefficients  of  m,  in  the  log.  of  2  first  found,  being  multi- 
plied by  this  number,  the  product  is  the  common  log.  of  2. 


INTRODUCTION.  25 

Hence,  the     common  log.  of  2  =  .3010299956. 

From  this  log.  we  may  find  the  logarithm  of  any  power  of  2 

by  multiplication  alone. 

Art.  18.  The  common  logarithm  of  any  prime  number 
may  be  readily  computed  when  that  of  the  next  inferior 
integral  number  is  known. 

Let  p  =  the  number  whose  log.  is  required ;  q  =  the  pre- 
ceding whole  number ;  R  =  2m  =  .8685889638  ;  and  putting 

p    _l  +  z 

n  j  v— <1        * 

we  find,  z  =  — : —  =  — ,— . 

p  +  q       p  +  q 

As  an  example,  let  the  log.  of  9  be  required ;  that  of  8,  the 
third  power  of  2,  being  known. 

In  this  case,  p  =  9,  q  =  8,  and  z  =  fo     Hence,  the 

log.  of  1  =  y7+  Jpp  +  ,_!_  +  >.  Jl^  =  .0511525224 ; 

in  which  A  is  the  preceding  term ;  B,  C  the  preceding  terms 
without  the  divisors,  3,  5. 

To  the  log.  of  f  add  the  log.  of  8,  or  three  times  the  log.  of 
2 ;  the  sum  .9542425093  =  log.  of  9 ; 

and  its  half,  or       .4771212546  ==  log.  of  3. 

From  these  logarithms,  the  logarithms  of  all  the  powers  of 
3,  and  of  all  the  products  of  2  and  3,  and  of  all  the  products 
of  their  powers,  may  be  obtained  by  multiplication  and 
addition. 

As  a  second  example,  let  the  log.  of  49  or  T2  be  required, 
the  log.  of  48  being  known  from  those  of  2  and  3.     Here, 

1  +  z 
p  =  49 ;  q  =  48 ;    , =  || ;  and  z  =  ^.     Hence, 


R 

97  '   *'(97Va 
4  c 


log.  of  i% \  =  —  +  J.T^r  =  .0089548426  ; 


26  PLANE  TRIGONOMETRY. 

to  this  add  the  log.  of  48 ;  and  the  sum  =  1.6901960797  is 
the  log.  of  49,  and  its  half  =  .8450980398  is  the  log.  of  7. 

Art.  19.  Although  the  methods  already  explained  are 
sufficient  to  enable  the  student  to  compute  the  logarithm  of 
any  given  number,  yet  there  are  other  expedients  for  abridg- 
ing the  labour  of  such  computations;  one  of  which  is  the 
following : 

Let  «,  b,  c,  be  three  equi- different  numbers,  whose  common 
difference  is  1 ;  so  that  a  —  b  —  1,  and  c  —  b  +  1 ;  then  ac  =* 
b2  —  1,  and  ac  +  1  =  62 ;  consequently, 

ac+l_  b* 

ac     ~~  ac' 


If  now  we  put  the  first  member  of  this  equation  in  place  of?/ 
1  + 
1  — 


1  +  z 
or  ^ in  the  general  equation,  (Art.  16,)  we  shall  have 


1 

z  = 


2ac  +  V 

a  quantity  which  will  converge  more  rapidly,  the  greater  a 
and  c  are.     Finding,  then,  the  log.  of 

ac  +  l 


we  have  the  log.  of 


ac 

ll 
ac 


If,  now,  the  logarithms  of  any  two  of  these  numbers  a,  b,  c, 
are  known,  the  log.  of  the  third  is  immediately  determined. 
For,  put 

A=log.  of  a ;  B=log.  of  b ;  C==log.  of  c ;  and  S=log.  of ; 

ac 

then,  since 

ac+l  _b*  ^ 
ac         ac1 


INTRODUCTION.  27 

it  follows  that  S  =  2B  —  A  —  C ; 

whence  either  A,  B,  or  C  being  required,  is  immediately  de- 
termined by  means  of  the  others. 

As  the  series  for  computing  the  log.  of 
ac+l 

converges  more  rapidly,  when  a,  &c.  are  large  numbers,  than 
when  they  are  small  ones,  the  labour  is  frequently  abridged 
by  computing  the  log.  of  a  power  or  multiple  of  the  number 
whose  log.  is  required. 

Let  the  log.  of  1 1  be  required,  those  above  computed  being 
considered  as  known. 

Here  we  may  take  a  —  98,  b  —  99,  and  c  =  100 ;  whence 
1  1 

Z~2ac+l~19601; 

and  S  =  log.  of  ^±^  =  j^.  =.0000443135, 

°  ac         19601 

the  other  terms  being  rejected,  because  they  do  not  affect  the 
result  short  of  the  fourteenth  decimal.  Now,  the  log.  of  98 
is  known  from  those  of  49  and  2,  and  the  log.  of  100  =  2,  the 
log.  of  (10)2.     Consequently,  in  the  equation 

S  =  2B  — A  — C; 
the  terms  are  all  known  except  B.     Therefore, 

B  =  i  (A  +  C  +  S). 


A  _    c    .3010299956    . 
1  1.6901960797    . 

.    log.  of     2. 
.    log.  of  49. 

C=      2. 

.    log.  of  100. 

S  =        .0000443135 

2)3.9912703888 

1.9956351944  -  . 

.     log.  of  99. 

But                   .9542425093    . 

.    log.  of  9. 

wherefore,      1.0413926851     . 

.    log.  of  11. 

28  PLANE  TRIGONOMETRY. 

Art.  20.  As  the  radix  of  the  common  system  is  10,  the 
log.  of  10  =  1,  the  log.  of  100  =  2,  the  log.  of  1000  =  3,  &c. ; 
hence  it  follows  that  the  log.of  any  number  less  than  10  con- 
sists wholly  of  decimals  ;  the  log.  of  a  number  which  is  more 
than  10,  but  less  than  100,  is  more  than  1,  but  less  than  2 ; 
the  log.  of  a  number  which  is  more  than  100,  but  less  than 
1000,  is  more  than  2,  but  less  than  3,  &c. ;  that  is,  if  the 
number  is  between  1  and  10,  the  integral  part  of  the  log.  is 
0 ;  if  the  number  is  between  10  and  100,  the  integral  part  of 
the  log.  is  1 ;  if  the  number  is  100  or  more,  but  less  than  1000, 
the  integral  part  of  the  log.  is  2.  This  integral  part  is  usually 
termed  the  index  of  the  logarithm. 

As  a  number,  when  multiplied  or  divided  by  any  power 
of  10,  is  still  indicated  by  the  same  significant  figures,  the 
position  of  the  decimal  point  only  being  changed  by  the  pro- 
cess ;  so  the  logarithm  of  a  number  being  increased  or  .dimi- 
nished by  adding  or  subtracting  the  log.  of  any  power  of  10, 
suffers  no  change  except  in  the  index  or  integral  part. 
Hence  we  readily  perceive  that  the  index  of  the  log.  will  be 
0,  1.  2,  or  3,  according  as  the  first  left-hand  significant  figure 
of  the  corresponding  natural  number  denotes  units,  tens, 
hundreds,  or  thousands. 

The  log.  of  1  being  0,  the  log.  of  a  proper  fraction  must  be 
negative ;  yet,  as  a  decimal  number  is  equivalent  to  an  inte- 
gral one  divided  by  some  power  of  10,  the  log.  of  a  decimal 
number  differs  in  nothing  but  the  index  from  the  log.  of  a 
whole  number  which  is  indicated  by  the  same  significant 
figures.  The  relation  between  the  logarithmic  index  and  the 
power  of  10  denoted  by  the  left-hand  digit  of  the  correspond- 
ing natural  number,  may  be  illustrated  by  arranging  the 
integral  logarithms  and  their  corresponding  natural  numbers 
in  adjacent  lines,  as  follows : 

3      2     10—1     —2  —3  —4      Log. 

ioooioo  io  iTV=.iTk=.oiToU-.oooiT4oo=-oooirj§! 

Here  it  is  evident  that  if  a  natural  number  falls  between  1 


INTRODUCTION.  29 

and  10,  its  log.  will  fall  between  0.  and  1.,  or  it  will  consist 
wholly  of  decimals.  If  the  number  is  between  .1  and  L,  the 
log.  will  be  between  — 1  and  0 ;  that  is,  the  index  of  the  log 
will  be  — 1.,  while  the  decimal  part  of  it  will  be  positive.  In 
like  manner,  when  the  natural  number  lies  between  .01  and 
.1,  the  index  of  the  log.  must  be  — 2,  and  the  decimal  part 
of  it  a  positive  quantity.  Hence  we  observe  that  when  the 
natural  number  consists  wholly  of  decimals,  the  logarithmic 
index  will  be  — 1.,  — 2.,  — 3.,  &c,  according  as  the  left-hand 
significant  figure  of  the  natural  number  denotes  tenths,  hun- 
dredths, thousandths,  &c. 

In  printing  tables  of  logarithms,  it  is  usual  to  omit  the 
index,  leaving  it  to  be  supplied  in  practice  upon  the  prin- 
ciples above  explained. 


SECTION  I. 


PLANE  TEIGONOMETRY. 

The  object  of  Plane  Trigonometry  is,  when  of  the  sides 
and  angles  of  a  plane  triangle  we  have  enough  given  to  limit 
it,  to  determine  the  parts  which  are  not  given. 

As  every  oblique  angled  triangle  may  be  divided  into  two 
right  angled  ones,  it  is  found  most  expedient  to  commence 
the  subject  by  examining  the  relations  and  properties  of  tri- 
angles of  the  latter  kind.  The  terms  of  the  science  are 
therefore  adapted  chiefly  to  right  angled  triangles. 


Definitions. 

Article  21.  Definition  1.  An 
arc  of  a  circle  is  any  part  of  the 
circumference,  usually  taken  less 
than  the  whole.  As  AB,  or  BHD. 

2.  The  chord  of  an  arc  is  a 
right  line  drawn  from  one  end  of 
the  arc  tQ  the  other.  Thus,  BE 
is  the  chord  of  the  arc  BAE,  or 
BDE. 

3.  The  sine  of  an  arc  is  a 
straight  line  drawn  from  one  ex- 

p  tremity  of  tKe  arc,  at  right  angles 

to  the  diameter,  which  passes  through  the  other  extremity. 

Thus,  AD  being  a  diameter  to  the  circle,  the  line  BF,  at  right 

angles  to  it,  is  the  sine,  or  right  sine,  of  the  arc  AB  or  DHB. 

4.  The  tangent  of  an  arc  is  the  right  line  which  touches 

(31) 


32  PLANE  TRIGONOMETRY. 

the  circle  at  one  extremity  of  the  arc,  and  extends  till  it 
meets  another  right  line,  which  is  drawn  from  the  centre 
through  the  other  extremity.  Thus  AG,  which  touches  the 
circle  at  A,  is  the  tangent  of  AB. 

5.  The  secant  of  an  arc  is  the  right  line  intercepted  be- 
tween the  centre  of  the  circle  and  the  extremity  of  the 
tangent.     Thus  CG  is  the  secant  of  the  arc  AB. 

6.  The  versed  sine  of  an  arc  is  the  part  of  the  diameter 
intercepted  between  one  end  of  the  arc,  and  the  sine  which 
passes  through  the  other  end.  Thus  AF  is  the  versed  sine 
of  AB,  and  DF  is  the  versed  sine  of  DHB. 

7.  The  part  by  which  an  arc  differs,  in  excess  or  defect, 
from  a  quadrant,  or  fourth  part  of  the  circumference,  is  called 
its  complement.  Thus,  the  arc  ABH  being  a  quadrant,  HB 
is  the  complement  of  AB  or  of  DHB. 

8.  The  cosine,  cotangent  or  cosecant  of  an  arc,  is  the  sine, 
tangent  or  secant  of  the  complement  of  that  arc.  Thus  BI, 
HK  and  CK,  the  sine,  tangent  and  secant  of  HB,  are  termed 
the  cosine,  cotangent  and  cosecant  of  AB. 

9.  What  an  arc  lacks  of  a  semicircle,  is  called  its  supple- 
ment.    Thus  BUD  is  the  supplement  of  AB. 

10.  The  circumference  of  every  circle  is  supposed  to  be 
divided  into  360  equal  parts,  called  degrees ;  each  degree 
into  60  equal  parts,  called  minutes',  each  minute  into  60 
equal  parts,  called  seconds,  &c.  Degrees,  minutes  and 
seconds  are  designated  thus,  °,  ',  ". 

11.  As  angles  at  the  centre  of  a  circle  have  to  each  other 
the  same  ratio  as  the  arcs  on  which  they  stand  (33.X3) ;  the 
latter  are  usually  termed  the  measures  of  the  former.  Hence 
an  angle  at  the  centre  of  a  circle  is  said  to  contain  as  many 
degrees,  minutes  and  seconds,  as  the  arc  which  subtends  it. 
The  sine,  tangent,*  &c,  of  an  arc,  is  also  called  the  sine,  tan- 
gent, &c.  of  the  angle  which  is  measured  by  the  arc.  Thus 
BF,  the  sine  of  AB,  is  called  the  sine  of  the  angle  ACB. 


SECTION  I. 


General  Properties  and  Relations  of  Arcs,  Sines,  Tan- 
gents,  $$c. 

Art.  22.  If  the  arcs  AH  and  DH  are  quadrants,  and 
therefore  equal,  it  follows  (33.6)  that  the  angles  ACH  and 
DCH  are  equal,  and  therefore  are  right  angles.  Hence  the 
angle  at  the  centre  of  a  circle,  subtended  by  a  quadrant,  is 
always  a  right  angle. 

Art.  23.  The  lines  AG  and  HK, 'which  touch  the  circle 
at  A  and  H,  are  respectively  at  right  angles  to  CA  and  CH 
(18.3);  hence  CAG  and  KHC  are  right  angled  triangles.  Now 
the  lines  AG  and  CH,  being  at  right  angles  to  AC,  are  pa- 
rallel to  each  other  (28.1)  ;  consequently,  the  alternate  angles 
IICK  and  CGA  are  equal ;  wherefore  the  triangles  CHK  and 
GAC  are  similar.  The  triangle  CFB  is  also  evidently  similar 
to  CAG;  and  CIB  to  CHK;  therefore  those  four  triangles 
are  similar  to  each  other.  Also,  the  figure  CFBI  being  a 
parallelogram,  CI  =  FB,  and  CF  =  BI.  From  these  trian- 
gles we  have  of  course  the  following  analogies : 

1.  As  CF  :  FB  : :  CA  :  AG,  or  as  cosine  :  sine  : :  radius  :  tang. 

2.  As  CF  :  CB  : :  CA  :  CG,  or  as  cosine  :  rad.  : :  radius  :  sec't. 

3.  As  CI  :  CB : :  CH :  CK,  or  as  sine  :  radius  : :  radius  :  cosec. 

4.  As  AG :  CA  : :  CH  :  HK,  or  as  tang. :  radius  : :  radius  :  cotan. 

5.  As  CG  :  AG : :  CB :  BF,  or  as  secant :  tang.  : :  radius  :  sine. 

In  the  algebraic  formula?  used  to  express  the  relations  of 
sines,  tangents,  &c,  it  is  most  convenient  to  assume  the  ra- 
dius =  1.  Making,  therefore,  this  assumption,  we  may  con- 
vert the  foregoing  analogies  into  the  following  equations. 

,      sine         ■  sine 

1.    — : —  =  tangent ;  sine  =  cosine. tangent ;  cosine  = . 

cosme,        -  *=*?  &  tan 

2. —  =  secant ;  cosine.secant  =  1 ;  cos= — -  =  — -  ■ 

cosine  •  sec't      Vl  +  tan2 

5 


34 


PLANE  TRIGONOMETRY. 


3.  - —  =  cosecant ;  sine.cosecant  ==  1. 
sine 


1 


_  1        cosine 

4.  —  =  cotan ;  tan.cotan  =  1 ;  cotan  =  - —  =  — : . 

tan  „*       tan        sine 

Hence,  taking  P  and  Q  any  arcs,  tan  P.  cotan  P  =  tan  Q. 
cotan  Q;  consequently  (16.6), 

tan  P  :  tan  Q  :  :  cotan  Q  :  cotan  P. 


5.    sine  == 


tangent         tangent 


secant        V  1  +  tan2 


Art.  24.  It  is  sometimes  necessary  to  attend  to  the  alge- 
braic signs  of  these  quantities,  particularly  when  they  are 
reduced  to  general  formulae. 

An  arc  estimated  in  one  di- 
rection is  considered  as  posi- 
tive, and  in  the  opposite  direc- 
tion as  negative.  The  same 
may  be  said  of  the  sines,  tan- 
gents, &c.  Thus  the  arc  AB, 
its  sine  FB,  cosine  CF,  tan- 
gent AG,  and  secant  CG,  are 
considered  as  positive ;  but, 
when  estimated  in  the  oppo- 
site direction,  they  are  consi- 
dered as  negative.  Now,  we 
readily  perceive  that  when 
the  arc  is  less  than  a  quadrant,  as  AB  is,  the  sine,  tan- 
gent, &c,  are  all  positive.  But  if  we  take  the  arc  more 
than  a  quadrant,  but  less  than  a  semicircle,  as  AL,  the  sine 
LM  is  still  positive,  but  the  cosine  CM  is  negative,  being 
measured  from  C  in  a  direction  opposite  to  CF.  The  tangent 
AP  and  secant  CP  are  also  negative  ;  the  former  being  drawn 
in  a  direction  opposite  to  AG,  and  the  latter  not  produced 
from  C  through  L,  the  extremity  of  the  arc,  but  in  the  oppo- 


SECTION   I.  35 

site  direction.  If  we  take  the  arc  more  than  a  semicircle, 
but  less  than  three  quadrants,  as  AHDN ;  the  sine  MN  be- 
comes negative,  the  cosine  CM  also  negative,  the  tangent  AG 
positive ;  but  the  secant  CG,  not  being  produced  through  N, 
but  in  the  opposite  direction,  is  negative.  If  we  take  the  arc 
more  than  three  quadrants,  but  less  than  four,  as  AHDE ; 
the  sine  EF  is  still  negative,  but  the  cosine  CF  and  the 
secant  CP  are  positive ;  the  secant  being  produced  from  the 
centre,  through  the  extremity  of  the  arc,  till  it  meets  the 
tangent ;  but  the  tangent  AP  is  negative. 

These  signs,  when  prefixed  to  the  several  quantities  in  the 
preceding  equations,  are  found  to  be  conformable  to  the 
algebraic  rules  for  the  adaptation  of  signs.  In  the  first 
quadrant, 

+  sine  11  1 

~ .—  =  +  tan;  — =  +  sec;- — —  =  +  cosec;  — —  =  -f  cot. 

+  cosin  +  cos  -f  sine  +tan 

In  the  second  quadrant, 

+  sine  11  1 

—  == — tan; = — sec;  - — : —  =  -f  cosec;  — - — = — cot. 

— cosin  — cos  +sine  — tan 

In  the  third  quadrant, 

—  sine  11  1 

— =  +  tan; = — sec;  — : — == — cosec;  — - — =+cot. 

— cosin  — cos  — sine  -ftan 

In  the  fourth  quadrant, 

—  sine  11  1 


— tan;— =+sec; : — = — cosec; — —  = — cot. 

+  cosin  -fcos  — sine  — tan 

Art.  25.  It  is  easily  perceived  that  the  sine,  tangent,  &c, 
of  a  given  arc  are  limited,  being  dependent  upon  the  length  of 
the  arc ;  but  the  sine,  tangent,  &c,  of  an  angle,  being  the 
sine,  tangent,  &c,  of  the  measuring  arc,  whatever  may  be 
the  radius  with  which  that  arc  is  described,  evidently  admit 
various  values.     Thus  EC,  HI,  MN,  which  are  the  sines  of 


16 


PLANE  TRIGONOMETRY. 


BC,  FI,  KN,  respectively,  are  also  the  sines  of  the  angle  at 
A.  The  lines  BL,  FO,  KP,  which  are  the  tangents  of  the 
same  arcs,  are  likewise  the  tangents  of  the  angle  at  A. 

Art.  26.  It  appears  from  cor.  to  15.4,  that  the  side  of  a 
regular  hexagon,  inscribed  in  a  circle,  is  equal  to  the  radius 
of  the  circle.  But  the  side  of  a  regular  hexagon,  inscribed 
in  a  circle,  subtends  an  arc  of  60°;  hence  the  chord  of  60°  is 
equal  to  the  radius  of  the  circle.  Again,  since  a  quadrant 
subtends  a  right  angle  at  the  centre  of  the  circle  (Art.  22), 
it  is  evident  that  the  sine  of  a  quadrant,  or  90°,  is  the  radius 
of  the  circle  (see  Fig.  p.  34) ;  thus  HC  the  sine  of  AH,  or 
ACH  is  the  radius  of  the  circle.  Further,  if  we  suppose  CG 
to  bisect  the  right  angle  ACH,  we  shall  have  CGA  (which  = 
HCG,  by  29.1)  =  ACG ;  whence  AG  =  CA  ;  that  is,  the  tan- 
gent of  45°,  or  half  a  right  angle,  =  the  radius.  Thus  it 
appears  that  the  chord  of  60°,  the  sine  of  90°,  and  the  tangent 
of  45°,  are  respectively  equal  to  the  radius  of  the  circle. 


Trigonometrical  Propositions. 

Art.  27.  The  sines  of  two  angles  adapted  to  any  radius 
have  to  each  other  the  same  ratio  as  the  sines  of  the  same 
angles  adapted  to  any  other  radius. 


SECTION    I. 


37 


Let  BAC  and  BAD  be 
two  angles,  whose  sines 
adapted  to  the  radius 
AC  or  AB,  are  EC  and 
FD;  while  the  sines  of 
the  same  angles  adapted 
to  the  radius  AG  or  AH, 
are  KH  and  LI. 

Since  the  angles  at  E, 
F,  K  and  L,  are  right 
ones,  it  is  evident  that 
the  triangles  AEC  and  AK.H  are  similar ;  as  are  also  AFD 
and  ALL     Consequently, 

As  AC  :  CE  : :  AH  :  HK ; 
and  alternately, 

AC  :  AH  : :  CE  :  HK. 
In  like  manner, 

As  AD  :  AI  : :  DF  :  IL ; 
wherefore,  CE  :  HK  : :  DF  :  IL; 

and  again  alternately, 

CE  :  DF  : :  HK  :  IL. 

Q.  E.  D. 

Cor.  If  we  substitute  the  word  tangent  or  secant  in  place 
of  sine,  the  proposition  will  still  be  true ;  and  the  demonstra- 
tion will  be  made  out  in  the  same  manner  by  drawing  tan- 
gents to  the  circles  at  B  and  G,  and  using  those  tangents,  or 
their  secants,  instead  of  the  sines. 

Art.  28.  In  any  right  angled  plane  triangle,  as  the  hy- 
pothenuse  is  to  the  perpendicular,  so  is  radius  to  the  sine  of 
the  angle  at  the  base ;  as  the  hypothenuse  is  to  the  base,  so 
is  radius  to  the  cosine  of  the  angle  at  the  base ;  and  as  the 
base  is  to  the  perpendicular,  so  is  radius  to  the  tangent  of 
the  angle  at  the  base. 


PLANE   TRIGONOMETRY. 


Let  ABC  be  a 
triangle,  right  an- 
gled at  B ;  from  A, 
with  the  radius 
AD,  describe  the 
arc  DE,  measur- 
ing the  angle  A ; 
through  E  and  D, 
draw  the  sine  EH 
and  tangent  DF. 
Then  the  triangles 


ABC,  ADF  and  AHE,  being  similar, 

As  AC  :  BC  : :  AE  :  HE ; 
As  AC  :  AB  ::  AE  :  AH; 
AB  :  BC  ::  AD  :  DF; 
As  AC  :  BC  : :  radius 
As  AC  :  AB  : :  radius 
As  AB  :  BC  : :  radius 


and 

that  is, 


and 


the  sine  of  A  ; 
the  cosine  of  A ; 
the  tangent  of  A. 


Q.  E.  D. 


Art.  29.  In  any  right  lined  triangle,  the  sides  have  to  each 
other  the  same  ratio  as  the  sines  of  the  opposite  angles. 

Let  ABC  be  a  trian- 
gle ;  make  AE  =  BC ; 
from  the  centres  B  and 
A,  with  the  radii  BC 
and  AE,  describe  the 
arcs  CG  and  EH ;  from 
C  and  E,  let  fall  on  AB 
c  (produced  if  necessary) 
the  perpendiculars  CD 
and  EF ;  these  perpen- 
diculars are  the  sines 
of  CG  and  EH,  or  of 
the  angles  B  and  A  to 


SECTION    I. 


39 


the   radius   BC  or  AE.      Now,  from  the  similar   triangles 
ACD,AEF; 

As  AC  :  CD  ::  AE  :  EF  (4.6); 

and  alternately, 

AC  :  AE  : :  CD  :  EF,  : :  sine  of  B  :  sine  of  A ; 

these  sines  being  suited  to  anv  radius  whatever  (Art.  27). 

Q.  E.  D. 

Art.  30.  In  any  right  lined  triangle,  the  sum  of  any  two 
sides  is,  to  their  difference,  as  the  tangent  of  half  the  sum  of 
the  angles,  opposite  to  those  sides,  to  the  tangent  of  half  their 
difference. 

Let  ABC  be  the  tri- 
angle; AC,  AB,  the 
sides.  From  the  centre 
A,  with  the  distance 
AC,  describe  the  circle 
DCEF;  meeting  AB, 
produced  in  D  and  E ; 
and  CB,  produced  in 
F ;  join  AF,  DC ;  and 
through  E  draw  EG 
parallel  to  BC,  meeting 
DC  produced  in  G. 
Then  it  is  evident  that  DB  is  the  sum,  and  BE  the  difference, 
of  AC  and  AB.  The  outward  angle  CAD  of  the  triangle 
ABC,  is  equal  to  the  two  inward  and  opposite  angles,  ABC 
and  ACB  (32.1).  But  AEC,  at  the  circumference,  is  equal 
to  half  the  angle  CAD  at  the  centre  (20.3) ;  that  is,  AEC  = 
half  the  sum  of  ABC  and  ACB.  Again,  AC  =  AF ;  there- 
fore, AFB  =  ACB  (5.1).     But, 

ABC  =  AFB  +  BAF  (32.1)  =  ACB  +  BAF ; 


40 


PLANE   TRIGONOMETRY. 


consequently,  BAF  =  the  difference  between  ABC  and  ACB ; 
and  therefore  ECF  =  half  that  difference  (20.3).  But  EG 
being  parallel  to  BC,  the  angle  CEG  =  ECF.  Furthermore, 
the  angle  DCE  in  a  semicircle  being  a  right  one  (31.3),  ECG 
is  also  a  right  angle.     Now,  because  BC  is  parallel  to  EG ; 

DB  :  BE  : :  DC  :  CG  (2.6). 

But  CD  is  the  tangent  of  CED,  and  CG  the  tangent  of  CEG, 
suited  to  the  radius  EC ;  and  these  tangents  have  to  each 
other  the  same  ratio  as  the  tangents  of  the  same  angles 
adapted  to  any  other  radius  (Art.  27).     Hence, 

AC  +  AB  :  AC  —  AB  : :  tang  of  }  (ABC  +  ACB)  :  tang  of 
J  (ABC  —  ACB). 

Q.  E.  D. 

Art.  31.  In  any  right  lined  triangle,  having  two  unequal 
sides ;  as  the  less  of  those  sides  is  to  the  greater,  so  is  radius 
to  the  tangent  of  an  angle ;  and  as  radius  is  to  the  tangent 
of  the  excess  of  that  angle  above  half  a  right  angle,  so  is  the 
tangent  of  half  the  sum  of  the  angles  opposite  to  those  sides, 
to  the  tangent  of  half  their  difference. 

B  Let  ABC  be  the  trian- 

gle ;  AB  the  less,  and  AC 
the  greater  side.  Draw 
AD  at  right  angles  to  AC, 
and  equal  to  AB  ;  cut  off 
AE,  also  =  AB ;  and  join 
DE  and  DC.  Then,  DAC 
being  a  right  angle, 

DA  :  AC  : :  rad  :  tangent 
of  ADC  (Art.  28). 

Now,  because  AE  =  AD,  the  angle  ADE  =  AED ;  hence 
each  of  those  angles  is  half  a  right  angle.    Since  the  triangles 


SECTION   I.  41 

ADE  and  ADC  have  the  angle  at  A  common,  the  angles 

ADC  +  ACD  =  ADE  +  AED  =  2ADE. 

Again,  since 

ADC  =  ADE  +  EDC  =  AED  +  EDC ; 

and         AED  =  ACD  +  EDC  (32.1) : 

it  follows  that      ADC  —  ACD  =  2EDC ; 

that  is,  ADE  is  half  the  sum,  and  EDC  half  the  difference, 
of  ADC  and  ACD.     Hence  (Art.  30), 

As  tan  of  ADE  :  tan  of  EDC  ::  AC  +  AD  :  AC  — AD  ::  tan 
{  (ABC  +  ACB)  :  tan  J  (ABC  —  ACB). 

But  the  tangent  of  ADE  =  radius  (Art.  26);  hence  the  above 
analogies  are  the  same  as  those  announced  at  the  beginning 
of  this  article.  A  Q.  E.  D. 

Art.  32.  In  any  plane  triangle,  as  the  base  is  to  the  sum 
of  the  sides,  so  is  the  difference  of  the  sides  to  twice  the  dis- 
tance between  the  middle  of  the  base  and  the  perpendicular 
falling  upon  it  from  the  vertex  of  the  triangle. 

Let  ABC  be  a  trian- 
gle, whose  base  is  AB. 
From  the  vertex  C,  with 
the  greater  side  AC,  de- 
scribe the  circle  AEGF, 
cutting  BC  produced  in 
E  and  F,  and  AB  pro- 
duced in  G;  join  AE, 
FG;  bisect  AB  in  H, 
and  draw  CD  at  right 
angles  to  AB.  Then, 
since  CD,  which  passes 
through  the  centre  of  the  circle,  cuts  AG  at  right  angles, 

6  D* 


PLANE   TRIGONOMETRY. 


that  is,       AB  :  AC  +  BC 


AD  =  DG  (3.3) ; 
or       AG  =  2AD  ; 
and    AB  =  2AH; 

therefore, 

BG  fe  2HD. 

Now,  the  angle  BAE  — 
BFG  ;  and  AEB  =  BGP 
(21.3);  consequently,  the 
triangles  ABE,  FBG,  are 
similar ;  wherefore, 
AB  :  BE  ::  BF  :  BG; 

AC  —  BC  :  2HD. 

Q.  E.  D. 


Art.  33.  If  the  half  difference  of  two  unequal  magnitudes 
be  added  to  the  half  sum,  the  result  is  the  greater  magnitude ; 
but  if  the  half  difference  be  subtracted  from  the  half  sum, 


the  result  is  the  less  magnitude. 


E 


B 


1  Let  AB  and  BC  de- 

c  note  any  two  unequal 
magnitudes,  whose  sum  is  AC  and  half  the  sum  AE  or  EC ; 
make  AD  =  BC ;  then  DE  =  EB,  the  half  difference.  Now, 
AB  =  AE  +  EB ;  and  BC  =  EC  —  EB.  Q.  E.  D. 

Art.  34.  When  the  sides  of  a  triangle  are  given,  we  have 
the  three  following  proportions  for  finding  either  of  the 
angles. 

Find  half  the  sum  of  the  three  sides,  and  from  that  half 
sum  subtract  the  sides  severally.     Then, 

1.  As  the  rectangle  of  the  half  sum,  and  the  excess  thereof 
above  the  side  opposite  the  proposed  angle,  is  to  the  rectan- 
gle of  the  other  two  remainders ;  so  is  the  square  of  radius 
to  the  square  of  the  tangent  of  half  the  angle. 

2.  As  the  rectangle  of  the  sides  containing  the  required 
angle  is   to  the  rectangle  of  the  excesses  of  the  half  sum 


SECTION   I. 


above  those  sides  respectively ;  so  is  the  square  of  radius  to 
the  square  of  the  sine  of  half  the  angle. 

3.  As  the  rectangle  of  the  sides  containing  the  required 
angle  is  to  the  rectangle  of  the  half  sum,  and  the  excess 
thereof  above  the  side  opposite  to  the  proposed  angle ;  so  is 
the  square  of  radius  to  the  square  of  the  cosine  of  half  the 
angle. 

Let  ABC  be  the  triangle ;  produce  AB,  AC,  to  II  and  L. 
Bisect  the  angles  BAC,  ABC  and  HBC,  by  the  lines  AG,  BG 
and  BK,  respectively;  and  let  AG,  BG,  meet  in  G.  Now, 
since  the  angle  CBH  is  greater  than  BAC  (16.1),  it  is  obvious 
that  HBK  is  greater  than  BAG;  and,  therefore,  (13.1,)  BAG 
-f  ABK  is  less  than  two  right  angles;  consequently  (cor.  29.1), 
BK  and  AG  produced  will  meet.  Let  them  meet  in  K;  and 
A  draw    KH,   KM,   EL, 

and  GD,  GF,  GE,  re- 
spectively, perpendicu- 
lar to  AB,  BC  and  AC. 
Then  it  is  obvious  (4.4) 
that  DG,  FG  and  EG 
are  equal ;  as  well  as 
KH,  KM  and  KL. 

Now,  in  the  triangles 
ADG,  AEG,  the  side 
AG  being  common,  and 
the  perpendiculars  DG, 
EG,  equal,  we  have 
(47.1)  AD  =  AE.  For  a  like  reason,  BD  =  BF,  CE  =  CF, 
BH  =  BM,  CL  =  CM,  and  AH  =  AL.  As  BH  =  BM,  and 
CM  =  CL,  it  follows  that 

AH  +  AL  =  BC  +  AB  +  AC. 

Hence  AH  or  AL  is  equal  to  half  the  sum  of  the  sides.     But 
that  half  sum  =  AD  +  BD  +  CF=AD  +  BC=BD  +  AC. 
Hence,  AH  — BC=AD; 


44  PLANE  TRIGONOMETRY. 

AH  —  AC  =  BD; 

AH  — AB  =  BH. 

Now,  the  angles  ABC 
and  CBH,  being  together 
equal  to  two  right  angles 
(13.1),  DBG  +  HBK  = 
one  right  angle  =  DBG  + 
L  BGD.  Consequently,  the 
triangles  DBG,  HBK,  are 
equiangular.  Also,  the 
triangles  ADG  and  AHK 
are  equiangular. 

Hence  (4.6) 

BD  :  DG  ::  HK  :  HB; 
wherefore  (16.6)       BD.HB  =  HK.DG. 
Also,  AD  :  DG  ::  AH  :  HK; 

consequently  (23.6), 

AD2  :  DG2  : :  AH.AD  :  HK.DG,  or  BD.HB. 
But  (Art.  28  and  22.6), 

AD2  :  DG2  : :  rad2  :  tan2  DAG  or  JBAC. 
Therefore, 

AH.AD  :  BD.HB  : :  rad2  :  tan2  JBAC. 
This  is  the  first  proportion. 

Again,  it  has  been  proved  that  CF  =  CE,  and  GF  =  GE, 
CG  being  common;  hence  (8.1)  the  angle  GCF  =  GCE; 
wherefore, 

GCA+GAC(=CGK)=JBCA+pAC=JHBC(32.1)=HBK. 

Consequently  (13.1),  AGC  —  ABK;  these  angles  being  the 
supplements  of  CGK,  HBK.  Also  the  angle  GAC  =  BAK. 
Therefore,  the  triangles  AGC  and  ABK  are  equiangular; 
whence  (4.6), 


SECTION   I.  45 

AB  :  AK  ::  AG  :  AC  .-.  AK.AG  =  AB.AC. 
Also  (4.6),  AG  :  GD  : :  AK  :  HK; 

wherefore  (23.6), 

AG2  :  GD2  : :  AK.AG  :  HK.DG  : :  AB.AC  :  BD.IIB. 
But  (Art.  28  and  22.6) 

AG2  :  GD2  : :  rad2  :  sin2  DAG  or  sin2  ^BAC. 
Hence,      AB.AC  :  BD.HB  :  rad2  :  sin2  JBAC ; 
which  is  proportion  2d. 
Further : 

AG  :  AD  : :  AK  :  AH  .-.  (23.6)  AG2  :  AD2  :  :  AK.AG  : 
AH. AD  ::  AB.AC  :  AH. AD. 

But  (Art.  28  and  22.6), 

AG2  :  AD2  : :  rad2  :  cos2  DAG  or  cos2  *  BAC. 

Consequently, 

AB.AC  :  AH.AD  : :  rad2  :  cos2  JBAC ; 

which  is  the  third  proportion. 


46 


PLANE  TRIGONOMETRY. 


SECTION  II. 

The  properties  of  plane  triangles,  which  are  explained  in 
the  preceding  section,  are  sufficient,  with  the  aid  of  the  usual 
auxiliary  tables,  to  enable  the  student  to  solve  all  the  com- 
mon cases  in  Plane  Trigonometry.  But  for  the  solution  of 
more  complex  problems,  and  particularly  for  the  purpose  of 
understanding  the  manner  in  which  the  trigonometrical 
tables  are  computed,  it  is  necessary  to  investigate  other 
theorems.  This  is  most  readily  effected  by  the  analytical 
method.  In  what  follows,  the  radius  to  which  the  sines, 
tangents,  &c,  are  adjusted,  is  always  taken  =  1.  But  it 
may  be  observed  that  whenever  it  is  required  to  apply  the 
results,  here  obtained,  to  the  case  where  the  radius  is  denoted 
by  any  other  number,  nothing  more  is  necessary  than  to 
change  all  the  trigonometrical  lines  in  the  same  ratio  in 
which  the  radius  is  changed. 


Article  35.  Let 
AB,    AC,   AD,  be 

three  equidifferent 
arcs,  whose  common 
difference  is  BC  or 
CD.  From  the  cen- 
tre O,  draw  OA, 
OC ;  from  B,  C,  D, 
draw  BE,  CF,  DG, 
at  right  angles  to 
OA ;  join  BD,  meet- 
ing OC  inn;  through 
B,  n,  draw  BH,  nm,  parallel  to  AO ;  and  np  parallel  to  CF. 
Then,  since  the  arc  BC  =  CD,  if  we  suppose  BO  and  DO 
joined,  those  arcs  will  subtend  equal  angles  at  O  (27.3). 


SECTION    II.  47 

Hence  (4.1)  Bn  =  Dn  ;  and  BnO  =  DnO;  consequently,  DnO 
is  a  right  angle.  Hence,  BE  =  sin  AB ;  CF  —  sin  AC  ;  DG 
=  sin  AD ;  Dn  =  sin  CD  or  BC ;  OE  =  cos  AB ;  OF  =  cos 
AC  ;  OG  ==  cos  AD  ;  On  =  cos  CD  or  BC.  Since  np  is  parallel 
to  CF,  and  nm  to  BH,  it  is  obvious  that  the  triangle  Onp  is 
similar  to  OCF;  and  Dnm  to  DBH;  and  as  DB  =  '2Dn,  it 
follows  that  BH  =  2nm ;  and  DH  =  2Dm.  Since  DnO  = 
mnp,  both  being  right  angles,  Dnm  =  Onp ;  and  the  angles  at 
m  and  p  are  right  ones ;  therefore  Dnm,  Onp,  are  similar  tri- 
angles. Of  course,  the  three  Dnm,  Onp,  OCF,  are  similar. 
Hence,  we  have  the  following  proportions : 

As  OC  :  CF  : :  On  :  np. 
As  OC  :  OF  : :  On  :  Op. 
As  OC  :  OF  : :  Dn  :  Dm. 
As  OC  :  CF  : :  Dn  :  nm. 

Taking  now  OC  =  1,  and  substituting  for  CF,  OF,  &c, 
sin  AC,  cos  AC,  &c,  these  proportions  furnish  the  following 
equations : 

np  =  sin  AC.  cos  CD.      (A) 

Op  ps  cos  AC.  cos  CD.      (B) 

Dm  a  cos  AC.  sin  CD.      (C) 

nm  ==  sin  AC.  sin  CD.      (D) 

From  equations  A  and  C, 

sin  AD(=np  +  Dm)=sin  AC.  cos  CD  +  cos  AC.  sin  CD.  (1) 
and 

sin  AB(=np  —  Dm)=sin  AC.  cos  CD  —  cos  AC.  sin  CD.  (2) 
From  equations  B  and  D, 

cos  AD(=Op  —  nm)  =  cos  AC.  cos  CD  —  sin  AC.  sin  CD.  (3) 
and 
cos  AB(=Op+w»i)=cos  AC.  cos  CD  +  sin  AC.  sin  CD.     (4) 


48  PLANE  TRIGONOMETRY. 

By  adding  equations  1,  2*; 

sin  AD  +  sin  AB  =  2  sin  AC.  cos  CD.  (5) 
By  subtracting, 

sin  AD  —  sin  AB  *  2  cos  AC.  sin  CD.  (6) 
B j  adding  equations  3,  4 ; 

cos  AB  +  cos  AD  =  2  cos  AC.  cos  CD.  (7) 
By  subtracting, 

cos  AB  —  cos  AD  =  2  sin  AC.  sin  CD.  (8) 


Art.  36.  From 
these  equations,  a 
number  of  others 
may  be  deduced. 

Suppose    AC  = 
CD  ;  and  put  AC= 
a ;    then   AD  =  2a, 
and  AB  =  0;   and 
Fp  G  U 

equation  1  becomes  sin  2(1=2  sin  a.  cos  a.  (1) 

equation  3,  cos  2a  =  cos2  a  —  sin2  a.  (2) 

equation  4,  cos  0=1  =  cos2  a  +  sin2  a, 

which  corresponds  with  47.1. 

From  the  equations  for  sin  2a  and  cos  2a,  it  is  manifest 
that  sin  a  =  2  sin  \d.  cos  \a.       (3) 


E 


cos  a  =  cos3  \t 


sin2  \a. 


and 

But     cos3  \a  =  1  —  sin3  \a.  .-.  cos  a  =  1 

From  the  last,       2  sin3  \a=\  —  cos  a. 


(4) 

-2sin*ia.      (5) 

(6) 


SECTION  II.  49 

In  equation  4,  substitute  for  sin3  \a  its  equal  1  —  cos2  \a, 
and  the  equation  becomes 

cos  a  =  2  cos3  %a  —  1.      (7) 

From  this  equation, 

2  cos3  \a  =  1  +  cos  a.      (8) 

By  Art.  23.1, 

sin  \a       2  sin  \a.  cos  \a       .       J  L. 

tan*a  =  ^  =  — ai^T1  =<e*3'8> 

(9) 


1  -f  cos  a  " 
By  Art.  23.4, 

cos  \a      2cosAa.  sinia     .      _    '       sin  a  ,_-. 

cotan  4a=  -^-f-  = o^Vi  '       =  (eq-3>  6)l •  (10) 

J       sin  fa  2sm~$a  vn       71  —  cos  a  x     ' 

sin3  ia       2sin2  \a     '       „  _1 — cos  a  .,"- 

tan"a  ^s?fe = S5s?fcr<,»  6'8>tt^w  <n> 

cotanHa=^.^L  =  |£2!;i_«=(eq.6)8)i±-c^.  (i2) 

2         sin3  ia         2  sin2  £a     v  n         '  1  —  cos  a  v     ' 


Art.  37.   Take   now  AC  =  a,  CD  =  b  ;  whence 

AD  =  a  +  & ; 
and  AB  =  a  —  b ; 

and  equations  1  and  2,  Art.  35,  become 

sin  (a  ±  5)  =  sin  a.  cos  b  =fc  cos  a.  sin  6.     (1) 
3  and  4  become, 

cos  (a  =fc  i)  =  cos  a.  cos  Z>  qp  sin  a.  sin  5.     (2) 
Now, 

.      ,    ..       sin(a  ±  Z/)      sin  a.  cos  b  ±  cos  a.  sin  b  ■ 

tan  (a  =fc  &)  =  — 7--^  = , 1 ■*-  =  (di 

v  '      cos(a±  0)      cos  a.  cos  0  zp  sin  a.  sin  0         v 


50  PLANE  TRIGONOMETRY. 

viding  numerator  and  denominator  by  cos  a.  cos  b,  and  using 

.      sin  k  tan  a  =b  tan  b 

tan  for  )  ^ — -.     (3) 

cos  '  1  qp  tan  a.  tan  o      x  ' 

Equations  5,  6,  7,  8,  also  become 

sin  (a  +  b)  +  sin  (a  —  b)  =  2  sin  a.  cos  6.  (4) 

sin  (a  +  ft)  —  sin  (a  —  5)  =  2  cos  a.  sin  &.  (5) 

cos  (a  —  b)  +  cos  (a  +  b)  =  2  cos  a.  cos  b.  (6) 

cos  (a  —  b)  —  cos  (a  +  b)  =  2  sin  a.  sin  5.  (7) 

By  Art.  23.1, 

sin  a      sin  b      sin  a.  cos  5  ±  cos  a.  sin  & 

tan  a  ±  tan  &  =  ± ,  = t = 

cos  a      cos  o  cos  a.  cos  6. 

(eq.i)!l^±l)      (8) 
x  n    'cos  a.  cos  6      x  ' 

By  Art.  23.4, 

cos  b      cos  a       sin  a.  cos  b  ±  cos  a.  sin  # 

COt  ft  ±  COt  a  =— r  ±  n = : : 7 '  = 

sin  b       sin  a  sin  a.  sin  b 

sin  a.  sin  b      w 

By  changing  our  notation,  other  equations  may  be  de- 
duced. 

Let  AD  =  a ;  AB  =  b  ;  then  AC  =  $($  +  &),  and  DC  or 
BC  =  ^(a —  b).  With  this  notation,  equations  5,  6,  7,  8,  Art. 
35,  become 

sin  a  +  sin  b  =  2  sin  J(<z  +  5).  cos  ^(a  —  b).  (10) 

sin  a  —  sin  b  =  2  cos  J(a  +  £).  sin  J(«  —  b).  (11) 

cos  6  +  cos  a  =  2  cos  ^(a  +  6).  cos  J(a  —  &).  (12) 

cos  b  —  cos  a  =  2  sin  £(a  +  5).  sin  |(a  —  &).  (13) 


SECTION   IL  51 

Also, 

U    jl  h\  -  sin-a(a  +  &)  _  2  sin  *(a  +  6).  cos  £(a  —  h)  _. 
tan  3(a  +  6)  -  cog  |^j  +  ftj  -  2  cos" i(a  +  ^  cos  |p  _  ^ 

/       in  io\  sin  a  +  sin  6 

(eq.  10,  12) T.  (14) 

v  _■  '  cos  a  -f  cos  6 

sin  |(«  —  ft)  _  2  sin  *(a  —  ft),  cos  *(a  +  ft)_ 
~V  '  ~~  cos  J(fl  —  ft)  —  2  cos  i(a  —  b).  cos  l(a  +  b)~ 

/        n     mx  sin  a  —  sin  ft 

(eq.  11,  12) ; r.  (15) 

v  T  '  cos  a  +  cos  b 

cos  i(<2  +  ft)      2  cos  Ua  +  ft),  sin  {{a  —  b) 

cotan  Ua  +  b)  =  — — 77-'  ,   ,x  =  « — : — 77 — — tt — 5 — 77 r\ 

ZK  '        sin  \{a  +  b)       2  sin  |(a  +  b).  sin  £(«  —  b) 

sin  a— sin  ft 

=  (eq.  11,  13) 7- .  (16) 

v  ^  '  cos  0  —  cos  a  N     ' 

_  cos  J  (a  —  ft)       2  cos  |(a  —  ft),  sin  *(a  +  ft) 
cotan  j(n  —  ft)  =-2HT(fl-.fc)   ~  2  sin  ^(a  — ft),  sin  J(a  +  6) 

,^   4«»   sin  a  +  sin  ft  ;4__ 

=  (eq.  10,  13)  - — -t .  (17) 

v  n  '  cos  0  —  cos  a  x     ' 

tan  Ua  —  ft)       .       ,  A  ,  rxsin  a  —  sin  ft 

f) — r  r(  =  (eq.  14, 15)- .j.    .    ,,        (18) 

tan  £(a  +  ft)       v  ^  'sin  a  +  sin  ft  x     ' 

tan|(«  —  &)       ,      1K  1mcosft  — cosa 


From  equations  15  and  16, 

cot  \{a  -f  ft)       cos  a  +  cos  ft 
tan  J(a  —  b)  —  cos  ft  —  cos  a' 


(20) 


The  following  is  an  analytical  investigation  of  the  rules, 
already  given  in  Art.  34,  for  finding  an  angle  of  a  plane  tri- 
angle., when  the  sides  are  known. 


X 


52  PLANE   TRIGONOMETRY. 

Let  ABC  be  the  triangle ;  and  put  the  angle  ABC  =  B ; 
the  side  AB  =  c;  AC  =  b;  BC  =  a;  BD  =  d;  the  line  CD 
being  at  right  angles  to  AB.     Then  (12,  13.2), 

c2  +  a2  =  &2±2a7; 

the  sign  +  being  used  when  B  is  acute,  as  in  Fig.  1 ;  and 
the  sign  —  when  B  is  obtuse,  as  in  Pig.  2. 


Fig.  2. 


Hence, 


db<*  = 


c3  +  a'  —  fr 


2c 


But  (Art.  28) 


—  =  cos  B ; 
a 


the  cos  B  being  positive  or  negative,  according  as  the  angle 
is  acute  or  obtuse  (Art.  24).     Consequently, 


cos  B  = 


c9  +  a2  —  b» 
2ac 


and 

1  +  cos  B  =  1  + 

4  (cor.  to  5.2) 


cs-fa3_&*  __  ca+2ac+as— ft*  _(c+a)3-&3 
2ac  2ac  2ac 

(c  +  a  +  b)  (c  +  a—  b) 
2ac 


SECTION  II.  53 

Now,  Art.  36,  Form.  8, 

1  -f  cos  B  =  2  cos3  JB ; 
wherefore, 

(c  +  a  +  b)  (c  +  a—b)       |(c>6>fe).   {(c  +  a—  b) 


cos3  AB  = 


4ac  ac 

c  +  a  +  b\  s.(s  —  b) 


=  (puUing  ,  =  —  2—  )— ; ■—  ;      (A) 
which  is  Rule  3,  Art.  34. 

c3  +  a*  —b*  , 

Again,  from  the  equation  cos  B  = ~ »  we  nave 

i       ;       c»  +  as— 63       2ac+62— c2— as 
1  —  cos  B  =  1  - 


2ac  2ac 

b*—(cs—2ac  +  az)  _  b»—(c—a)a 
Qac  2ac 


But,  Art.  36,  Form.  6, 

1  —  cos  B  =  2  sina  JB ; 

consequently, 

63—  (c— a)*       l(b  +  c—a).  {(b  +  a—c) 


sin3  ^B  = 

iac  ac 

(B) 


ac 
which  is  Rule  2,  Art.  34. 

Since  tan  =  -^  (Art.  23) ;  tan3  B  =  — -^  =  (by  eq. 
cosine  v  '  cos8B       v  J      ' 

,  -r^  (s  —  a)  .(s  —  c) 
A  and  B)  J /  *  ■. x  -  '; 

'       s  .  (s  —  b) 

which  is  Rule  1,  Art.  34. 

E* 


PLANE  TRIGONOMETRY. 

Art.  38.   Let  ABC 

be  a  semicircle ;  ADC 
its  diameter;  D  the 
centre ;  AB  the  chord 
of  an  arc;  AE,  BG, 
lines  touching  the  cir- 
Iji  ~^c     cle  in  A  and   B,  and 

meeting  in  G.     Join  DG,  DB,  CB ;  and  produce  DB,  CB,  to 

meet  AE  in  E  and  F. 

Now,  the  angle  ABC  in  a  semicircle  being  a  right  one 
(31.3),  the  adjacent  angle  ABF  is  also  a  right  one  (13.1). 
Again,  since  AG  and  BG  touch  the  circle,  each  of  the  angles 
GAB,  GBA  ==  the  angle  ACB  in  the  alternate  segment  (32.3) ; 
hence  GAB  =  GBA,  and  GB  ==  GA  (6.1).  Furthermore, 
since  AFB  +  FAB  =  ABC  (32.1)  =  ABF  (31.3  and  13.1)  =? 
GBF  +  GBA ;  it  follows  that  GFB  =  GBF ;  whence  GB  = 
GF  ;  and  AG  +  GB  =  AF.  But  the  triangles  ADG,  BDG, 
being  evidently  equal,  the  line  AG  is  the  tangent  of  half 
the  arc  intercepted  between  A  and  B ;  hence  AF  =s  twice 
the  tangent  of  that  half  arc.  The  chord  AB  is  also  twice 
the  sine  of  the  same  half  arc.  Now,  the  triangles  ACB, 
FAB,  being  right  angled  at  B,  and  having  the  angle  ACB  = 
FAB,  are  similar ;  whence  AC  :  CB  : :  AF  :  AB  : :  AG  : 
JAB  : :  tan  of  ^  arc  :  sine  of  the  same  half  arc. 

Let,  now,  the  point  B  move  along  the  arc  towards  A ;  the 
lines  which  pass  through  B  moving  with  it :  then,  as  the 
angle  at  C  diminishes,  the  line  CB  must  approximate  to  AC, 
and  ultimately  become  equal  to  it.  Consequently,  the  ratio 
of  AF  to  AB,  or  of  the  tangent  to  the  sine  of  half  the  arc 
between  A  and  B,  is  ultimately  a  ratio  of  equality. 

Again.  The  angle  ADG  =  JADB  =  ACB  (21.3) ;  con- 
"sequently,  DG  is  parallel  to  CF  (28.1) ;  and  therefore, 

EB  :  BD  : :  EF  :  FG  (2.6) ; 


SECTION   II. 


or,  doubling  the  consequents, 

EB  :  AC  : :  EF  :  FA ; 
whence,  by  composition  (18.5), 

EB  +  AC  :  AC  : :  AE  :  AF. 
But  AC  :  CB  : :  AF  :  AB  (4.6) ; 

therefore,  ex  equali, 

AC  +  BE  :  CB  : :  AE  :  AB  (22.5). 

But  as  the  point  B  approaches  A,  the  line  BE  decreases 
and  ultimately  vanishes.  The  angle  at  C  ultimately  vanish- 
ing, 4he  line  CB  becomes  finally  equal  to  AC.  Hence  the 
ratio  of  AC  +  BE  to  CB,  and  consequently  of  the  tangent 
AE  to  the  chord  AB,  becomes  ultimately  a  ratio  of  equality. 

Hence  it  is  manifest  that  the  ultimate  ratio  of  the  tangent 
of  an  evanescent  arc  to  its  sine,  or  to  its  chord,  is  a  ratio  of 
equality. 

B  Let  AEB  be   a  circular  arc, 

whose  centre  is  C  ;•  AD,  BD,  two 
right  lines  touching  the  circle  in 
A  and  B.  Join  CD,  AB;  and  let 
CD  cut  the  arc  in  E,  and  the 
chord  AB  in  F.  Through  E, 
draw  GH  parallel  to  AB ;  and 
join  AE,  BE.  Then,  from  what 
is  above  proved,  AD  =  BD ;  the 
c  angle  ACF  =  BCF ;  and  conse- 
quently (4.1),  AFC  =  BFC;  whence  GIT,  being  parallel  to 
AB,  and  therefore  at  right  angles  to  CE,  touches  the  circle 
in  E.  Also,  AD  is  the  tangent  and  AF  the  sine  of  the 
arc  AE. 

Since  AFE  is  a  right  angle,  the  angles  AEF  and  ADF  are 
each  less  than  a  right  angle  (17.1).  But  AED  =  AFE  + 
EAF  (32.1)  is  greater  than  a  right  angle.  Hence,  AD  is 
greater  than  AE,  and  AE  than  AF  (19.1).     That  is,  the  tan- 


56 


PLANE   TRIGONOMETRY. 


gent  is  greater  than  the  chord,  and  the  chord  than  the  sine. 
Again,  since  GD  +  DH  are  greater  than  GH  (20.1),  it  is 
obvious  that  AD  +  DB  must  be  greater  than  AG  +  GH  + 
HB.  But  AG  +  GE  being  greater  than  AE  (20.1),  and  EH 
+  HB  than  EB ;  AG  +  GH  -f  HB  must  be  greater  than  AE 
+  EB.     Also  AE  +  EB  are  greater  than  AB. 

If  we  were  to  join  CG,  CH,  and  draw  the  tangents  and 
chords  to  the  intercepted  arcs,  we  might  demonstrate,  in  the 
same  manner,  that  the  sum  of  the  tangents  thus  drawn  would 
be  less  than  AG  +  GH  +  HB,  and  the  sum  of  the  chords 
greater  than  AE  +  EB.  By  continued  bisections,  we  thus 
find  the  sum  of  the  tangents  continually  decreasing,  and  the 
sum  of  the  chords  always  increasing.  But  the  tangents  and 
chords  are,  by  this  process,  brought  to  approximate  still 
more  and  more  nearly  to  the  circular  arc  which  lies  between 
them.  Hence  we  infer  that  when,  by  the  evanescence  of  the 
arc,  the  ratio  of  the  tangent  to  the  chord  or  sine  becomes  a 
ratio  of  equality,  the  ratio  of  the  arc  itself  to  the  tangent, 
chord  or  sine,  is  a  ratio  of  equality. 

D 


E  Fp 

tial  Calculus. 
Let  AB  = 
OC  being  =  1. 


Art.  39.  As  near- 
ly all  trigonometri- 
cal calculations  are 
usually  performed 
by  means  of  auxili- 
ary tables,  it  be- 
comes necessary  to 
explain  the  nature 
and  origin  of  those 
tables.  This  is  most 
expeditiously  effect- 
ed by  the  Differen- 


BE  =  y;  OE  =  x;  BCD  =  h;  the  radius 
Then,  as  proved  in  Art.  35,  equation  C, 
Dm  =  cos  (z  +  jjk).  sin  \h ; 
whence,  sin  (z  +  h)  =  sin  %  +  2  cos  (z  +  ^).sin  \h ; 


SECTION   II.  57 

consequently, 

sin  (z  +  h)  —  sin  z  _  2  cos  (z  +  *A).sin  J  A      cos(z-f-|A).sin|Zt 
A  h  \h 

But  (Art.  38)  when  h  becomes  evanescent,  the  ultimate  ratio 
of  }Ji  to  the  sin  \h  is  a  ratio  of  equality.  Also  the  ultimate 
ratio  of  cos  (z  +  \K)  to  cos  z  is  a  ratio  of  equality.     Hence, 

e/.sinz       /dy\ 

—fa     =te-)  =  cosz  =  *; 

consequently,  dy  =  xdz. 

Again,  x2  -f  yz  =  1 ;  whence,  2;«£c  -f  2*/c?!/  =  0;  or 

— ydy      — yxdz 

dx  =  —2-2-  =  — - =—ydz. 

x  x  J 

Now,  let  t  =  tan  z  =— ;  then 

x 


y        xdy  —  ydx    '  xldz  +  y*d% 

x  x4 


„^_£~   m  S=-«f  ^    (since  f+fim  1) 


Z  =  (]  +  *2)  <fe,  because -2  =  1  +  t2  (Art.  23,  2.) 


x- 


Art.  40.  To  find  the  length  of  an  arc  z  in  terms  of  its  tan- 
gent, t.  No  formula  has  been  discovered  for  expressing  an 
arc  in  finite  terms  of  its  tangent ;  recourse  must  therefore  be 
had  to  infinite  series. 

Let,  then,  z  =  Atn  +  B*p  +  Cf  +  Df,  &c,  in  which  the 
exponents  and  coefficients  are  unknowrn.  As  the  tangent  of 
an  arc  becomes  nothing  at  the  same  time  the  arc  itself 
vanishes,  the  series  here  assumed  must  express  the  arc  z,  if  it 
can  be  expressed  at  all  in  terms  of  t. 

From  this  equation, 

■^=n  At*-1  +  pBt^"1  +  qCt*-1  +  rBv-1  +  &c. 
8 


58  PLANE   TRIGONOMETRY. 

But  (Art.  39), 

dz 
Comparing  these  values  of  77;  n  —  1=0,  p  —  1  =  2,  q  —  1 

=  4,  r  —  1  =  6,  &c. ;  and  nA.  =  1 ;  pB  =  —  1 ;  qC  =  1 ;  ?*D 
=  —  1,  &c. ;  from  which  n  =  1;  p  =  3;  q  =  5, &c.  ;  A  =  1 ; 
B  =  —  ^ ;  C  =  I;  D  =  —  \,  &c. ;  which  values,  substituted 
in  the  primitive  equation,  give 

z  =  t — jf  +  j** — *f + jf  +  &c. 

The  length  of  the  arc  being  thus  found  in  terms  of  the 
tangent,  the  next  object  is  to  find  a  value  of  t  which  will 
converge  rapidly  in  this  series,  and  at  the  same  time  corre- 
spond to  a  known  part  of  the  circumference  of  a  circle. 
One  of  the  most  convenient  methods  yet  discovered  of  effect- 
ing this  object,  is  the  following : 

Let  a  and  b  denote  two  circular  arcs ;  such  that  tan  a=\, 
and  tan  b  =  J;  then  (Art.  37,  Form.  3), 


tan  {a  +  b)  = 

=  1. 

ow  (Art.  26)  1 

the  tangent  of  45 

'  =  i. 

Hence,  a  +  b  = 

=  45°. 

Now  a  - 

1 
=  2 

11        11 

1 1 

&c.= 

:. 463647609001 

and      b  =  -£  —  i*-^  +  y.  35—  yyp  &c.=.321750554397 


a  +  b  =  (45°)  = 785398163398 

From  which  we  find  the  semicircumference==3.141592653592; 
and  this  is  the  whole  circumference  very  nearly  true  to  twelve 
decimals,  when  the  diameter  is  =  1. 

Art.  41.  The  length  of  an  arc  of  45°  being  thus  deter- 
mined, the  length  of  an  arc  of  1°,  or  of  any  fraction  of  it,  is 


SECTION   II.  59 

readily  found.  But,  from  this  datum,  to  find  a  general  for- 
mula for  the  sine  and  cosine,  requires  the  aid  of  the  Differen- 
tial Calculus.  And  no  method  has  been  discovered  to  denote 
the  sine  or  cosine  in  terms  of  the  arc,  without  recourse  to 
infinite  series. 

Let  z  =  an  arc ;  y  =  sin  z ;  x  =  cos  z ;  and  assume 

x  =  1  +  Azn  +  Bzp  +  Czq  +  Dzr  +  &c. ; 

in  which  the  first  term  of  the  series  is  taken  =  1 ;  because, 
when  z  =  0,  x  =  1.     From  this  equation, 

dx 

-^  =  nAzn~l  +  pBz»-1  +  qCz*-1  +  rT>zr~l  -f  &c. 

But  (Art.  39), 

dx 

Therefore, 

y  =  —  nAz""1  —  pBzP"1  —  qCz*-1  —  rDz'"1  —  &c. ; 

and 

du 

fz=  —  (n  —  1).»  Az"-2  —  (p  —  l).pBzP-2  —  (7  —  l).?Cz<-» 

—  (r— l).rDzr~2  — &c. 

But  (Art.  39), 

-|  =  x  =  1  +  Azn  +  Bzp  +  Czq  +  &c. 

Comparing  the  corresponding  terms  of  these  series, 

n  —  2  =  0;  p  —  2=ti;  q  —  2  =  p;  r— -2  =  q,&e. 
(to  —  l).nA  =  —  1;  (/>  —  l).pB  =  —  A  ;  &c. 

Whence        n  =  2;  p  =  4;  ?  =  6;  r  =  8;  &c. 

A-    _Lr_J_    p_  1         n_         X         * 

A~      2'  *~  2.3.4'  U 2^Z^6;  U_2.3.4.5.6.7.8&c* 

Consequently, 


60  PLANE    TRIGONOMETRY. 


x  =  l  ~  1  +  2^4-3X47^6  +  &C-  (A) 

*  =  Z  ~is  +  2^475-2.3.416.7   +  &C>       (B) 


To  exemplify  these  series,  let  the  sine  and  cosine  of  1°  be 
required.     In  that  case, 

*=.017453292520;^=  .000152308710 ;  ^=.000000886080; 
m  .000000003866;  *&-==  .000000000013. 


2.3.4  — ^vyw'  2.3.4.5 

Substitute  these  values  in  series  A  and  B.  Whence  x  = 
.999847695156;  and  y  =  .017452406453. 

z*     z3 
If,  instead  of  these  values  of  z,-^,  s~q>  &c->  we  substitute 

in  the  series  A  and  B,  \  of  the  first,  \  of  the  second,  J  of  the 
third,  &c,  we  shall  obtain  the  cosine  and  sine  of  30' ;  and 
from  these  results  we  may,  by  a  similar  process,  find  the 
cosine  and  sine  of  15'. 

Thus,  cos  of 

30'=  1— .000038077177 + .000000000242= .999961923095 ; 
and  sin  of 

30'=.00872664626O—  .000000 11 0760=. 008726535500; 
Also,  cos  of 

15'= 1— .000009519299  +  .000000000015=  .999990480728 ; 
and  sin  of 

15'=. 004363323130— .000000013845=. 004363309285. 

z2 
If,  instead  of  the  values  of  z,  -~-,  &c,  first  found,  we  take 

e*3  of  the  first,  gg1^  of  the  second,  &c,  and  substitute  them 
in  equations  A  and  B,  we  shall   have  the  cosine  and  sine 

of  r. 


SECTION   II.  01 

Thus,  cos  of  r=l.— .000000042308 =.999999957692; 
and  sin  of 

1'= .000290888209— .000000000004=  .000290888205. 

When  the  sines  and  cosines  of  two  arcs  are  known,  the 
sine  and  cosine  of  their  sum,  or  difference,  are  readily  found 
from  equations  1  and  2,  Art.  37.  Or  the  sine  and  cosine  of 
any  arc,  v,  being  found,  the  sine  and  cosine  of  2u,  3v,  4v,  may 
be  determined  in  the  following  manner : 

From  Art.  37,  Form.  4  and  7,  we  have,  by  transposition, 

sin  (a  +  b)  =  2  sin  a.  cos  b  —  sin  (a  —  b)  (C) 

cos  (a  -f  b)  =  cos  (a  —  b) —  2  sin  a.  sin  b  (D) 

Taking,  then,  b  =  any  arc  v ;  and  a  successively  =  v3  2v, 
3v,  &c.  ;  these  equations  become 

sin  2v  =  2  sin  v.  cos  v ;  -    V    C*-> 

cos  2v  =  1  —  2  sin2  v ; 

sin  3u  =  2  sin  2v.  cos  v  —  sin  v ; 

cos  3v  =  cos  v  —  2  sin  2p.  sin  i; ; 

sin  4u  =  2  sin  3d.  cos  u  —  sin  2u  ; 

cos  4u  ==  cos  2v  —  2  sin  3v.  sin  u. 

Hence,  the  sine  and  cosine  of  v  being  known,  the  sine  and 
cosine  of  any  multiple  of  v  may  be  found. 

Art.  42.  The  sine  of  an  arc,  being  half  the  chord  of  twice 
the  arc,  and  the  chord  of  60°  =  1,  (Art.  26,)  the  sine  of  30° 
=  I ;  consequently,  if  we  take  a  =  30°,  in  the  equations  C 
and  D  of  the  last  article,  we  shall  hnve 

sin  (30°  +  b)  =  cos  b  —  sin  (30°  —  b)  ; 

cos  (30°  +  b)  =  cos  (30°—  b)  —  sin  b. 

If,  then,  the  sine  and  cosine  of  every  degree  and  minute, 
as  far  as  30°,  were  computed   by  the  preceding  methods, 

F 


t 


lA^s 


62  PLANE  TRIGONOMETRY. 

these  equations  furnisn  a  mode  of  computing  the  sines  and 
cosines  of  the  remaining  arcs  by  subtraction  only. 

To  find  the  sine  and  cosine  of  31°, 

cos  29°  =  cos  30°.  cos  1°+  sin  30°.  sin  1°  =    .874619707108 

sin  1°  =    .017452406453 


cos  31°=    .857167300655 

cos    1°=    .999847695156 
sin  29°  =  sin  30°.  cos  1°  —  cos  30°.  sin  1°  =    .484809620238 

cos.  31°  =    .515038074918 

The  sines,  computed  as  above  explained,  and  arranged  in 
a  table,  constitute  a  table  of  natural  sines.  Those  sines,  as 
put  down  in  the  tables,  are  seldom  extended  to  more  than 
seven  decimals,  and  frequently  not  even  so  far ;  but,  in  com- 
puting such  tables,  it  is  necessary  to  extend  the  sines  of  the 
primary  arcs  considerably  further  than  the  number  of  deci- 
mals intended  to  be  retained,  in  order  to  render  the  numerous 
deductions  from  them  sufficiently  correct. 

The  tangents  may  be  found  from  the  sines  and  cosines,  by 

sine 

simple  division;  for  tan  = —  (Art.  23).     The  secants  are 

1  cosine  v  ' 

also  deduced  from  the  cosines ;  for  secant  = 


cosine 


Art.  43.  The  tables  of  sines,  tangents,  &c,  which  are 
commonly  used  in  trigonometrical  calculations,  are  loga- 
rithmic, and  are  easily  deduced  from  a  table  of  logarithms 
and  of  natural  sines.  But  it  may  be  observed  that  the  sine? 
computed  to  a  radius  1,  are  all  decimals  except  the  sine  of 
90°.  Hence,  the  logarithms  of  those  sines,  if  the  common 
logarithms  are  used,  must  all  have  negative  indices,  excep 
the  sine  90°,  whe.se  logarithm  is  0.  To  avoid  this  inconve 
nience  the  decimal  point  in  each  of  the  sines  is  removed  tec 


SECTION   II.  63 

places  towards  the  right,  which  is  equivalent  to  finding  the 
sines  to  a  radius  of  10000000000.  The  logarithm  of  this 
number  is  10;  and  the  sine  of  1"  computed  to  this  radius  is 
48481.37,  whose  log.  =4.6855749.  From  which  it  appears 
that  all  arcs  or  angles  which  can  occur  in  practice,  have 
their  logarithmic  sines  positive. 

The  sines  computed  according  to  the  preceding  articles 
have  the  decimal  point,  in  each  case,  removed  ten  places  to 
the  right ;  the  logarithms  of  the  results  are  then  taken  from 
a  table  of  logarithms,  and  arranged  in  a  table.  This  com- 
poses a  table  of  logarithmic  or  artificial  sines.     Then,  since 

cos  :  sine  : ;  rad  :  tang  (Art.  23) ; 

the  index  of  the  logarithmic  sine  being  increased  by  10,  and 
the  logarithmic  cosine  subtracted,  the  remainder  will  be  the 
logarithmic  tangent.     And,  since 

cos  :  rad  : :  rad  :  secant  (Art.  23) ; 

if  we  subtract  the  logarithmic  cosine  from  20  (twice  the  log. 
of  radius),  the  remainder  will  be  the  secant.     Again, 

tang  :  rad  ::  rad  :  cotan; 

consequently,  the  logarithmic  tangent  of  an  arc,  being  sub- 
tracted from  20,  will  leave  the  logarithmic  cotangent. 

In  trigonometrical  calculations  where  logarithms  are  used, 
it  is  most  convenient  to  take  the  arithmetical  complements 
of  the  logarithms  which  are  to  be  subtracted,  (that  is,  what 
those  logarithms  want  of  10  or  20,)  and  add  them  with  the 
other  additive  logarithms,  rejecting  as  many  tens  or  twen- 
ties from  the  result  as  there  are  complements  used.  When 
the  subtractive  numbers  are  logarithmic  sines,  tangents  or 
secants,  the  arithmetical  complements  can  be  taken  imme- 
diately from  the  table ;  for  the  cosecant  is  the  arithmetical 
complement  of  the  sine ;  the  cotangent  of  the  tangent ;  and 
the  cosine  of  the  secant.  All  this  is  manifest  from  the  na- 
are  of  logarithms,  and  the  analogies  in  Art.  23. 


64 


PLANE  TRIGONOMETRY. 


Art.  44.  A  few  trigonometrical  problems  will  now  be 
given  to  exercise  the  preceding  rules. 

1.  Given,  AB  35,  AC  30, 
BC  25,  the  three  sides  of  a 
triangle ;  to  find  ihe  distances 
from  the  several  angles  to  a 
point  E  within  the  triangle, 
such  that  the  angles  AEB, 
AEC  and  BEC  shall  be  equal 
to  each  other. 

Construction.  On  AB,  one 
of  the  sides,  describe  the 
equilateral  triangle  ABD ;  and 
about  that  triangle  describe  a 
circle;  join  DC,  cutting  the 
circle  in  E ;  then  is  E  the  point  required. 

Join  AE,  BE ;  then,  since  the  angles  of  the  triangle  ABD 
are  all  equal  (cor.  to  5.1),  each  of  them  contains  60°,  or  ^  of 
two.right  angles  (32.1).  But  AED= ABD ;  and  BED=BAB 
(21.3) :  therefore  AED  =  60°,  and  AEC  =  120°.  Also  BEYJ 
=  60°,  and  BEC  =  120°. 

Calculation.  With  the  three  sides,  the  angle  BAC  may  Vf 
found  by  Art.  34,  Rule  2. 

AB  35  AC  8.4559320 
AC  30  AC  8.5228787 
BC        25 


sum 


90 


J  sum   45 
J  sum—  AB     10    log.  1. 
i  sum  —  AC     15      "    1.1760913 

2)19.1549020 


sinpAC   22°12J'      9.5774510 


SECTION   II.  65 

BAG    44°  25' 
BAD  60° 


DAC  104°  25 


£DAC   52°12J' 
ACD  +  ADC"^4, 

36 

Then,  by  Art.  30, 

As 
is  to 


So  is  tan 
to       tan 


AD  +  AC 

65        AC  84870866 

AD  — AC 

5                 .6989700 

ACD  +  ADC 

2 

37°  47£'      9.8895519 

ACD  — ADC 

2 

3°  24J'      8.7756085 

ACD 

41°  12' 

ADC) 
ABES 

34°  23' 

CAE 

18°  48=  AED— -ACE. 

EAB 

25°  37=  CAB  — CAE. 

From  Art.  29, 

As    sinAEB  120°  AC     .0624694 

is  to  sin  ABE  34°  23'  9.7518385 

So  is      AB  35  1.5440680 


to           AE  22.823  1.3583759 

As    sinAEB  120°  AC     .0624694 

istosinBAE  25°  37'  9.6358335 

So  is       AB  35  1.5440680 


to  BE       17.473  1.2423709 

9  F* 


G8  PLANE   TRIGONOMETRY. 

As     sin  AEC     120°        AC     .0624694 
is  to  sin  CAE     18°  48'  9.5082141 

So  is       AC       30  1.4771213 


to 


CE       11.164 


1.0478048 


2.  Given,  the  vertical  angle 
ACB  70°,  the  segments  into 
which  the  base  is  divided  by 
the  line  CD  bisecting  the  ver- 
tical angle,  viz.  AD  30,  and  DB 
20,  to  determine  the  angles 
and  sides  of  the  triangle,  and 
the  line  which  bisects  the  ver- 
tical angle. 


Construction.  Bisect  the  base  AB  in  E ;  through  E  draw 
FEG  at  right  angles  to  AB ;  make  the  angle  EAF  =  com- 
plement of  the  vertical  angle,  above  AB  when  that  angle  is 
acute,  but  below  when  it  is  obtuse.  From  the  centre  F, 
where  the  line  AF  meets  the  perpendicular,  describe  a  circle 
passing  through  A,  and  cutting  FEG  in  G;  join  GD,  and 
produce  it  to  cut  the  circle  in  C ;  join  CA,  CB ;  and  ACB 
will  be  the  triangle  required. 

Since  AE  =  BE,  and  the  angles  at  E  are  right  ones,  the 
line  AF  =  BF  (4.1) ;  consequently,  the  circle  must  pass 
through  B.  The  angle  AFE  being  equal  to  BFE,  the  arc 
AG  =  BG  (26.3) ;  consequently,  ACG  =  BCG  (27.3).  Also, 
the  angle  AFE  =  twice  ACG  (20.3)  =  ACB ;  therefore,  ACB 
is  the  complement  of  EAF. 

Calculation.  In  the  right  angled  triangle  EAF  we  have, 
besides  the  right  angle,  the  side  AE  =  25,  and  the  angle 
EAF  =  20° ;  from  which  we  find,  by  Art.  28,  AF  =  26.604, 


SECTION  II. 


67 


and  EF  =  9.099 ;  whence  EG  =  17.505 ;  then,  in  the  right 
angled  triangle  GED,  we  have  EG ;  and  ED  =  5 ;  from 
which  we  find  the  angle  EDG  or  CDB  =  74°  3' ;  then 

CAD  =  CDB  — ACD  (32.1)  =  39°  3'. 

In  the  triangle  ADC  we  then  have  AD  =  30,  and  all  the 
angles,  from  which,  by  Art.  29,  we  find  AC  =  50.29,  and 
DC  =  32.951.     Lastly,  from  3.6,  we  have 

AD  :  DB  : \  AC  :  BC  =  33.527. 


3.  Given,  the  base  AB  70; 
the  vertical  angle  ACB  75°; 
and  the  ratio  of  the  sides,  viz., 
AC  :  BC  : :  4  :  3,  to  deter- 
mine the  rest. 

Divide  AB  in  D,  so  that 
AD  :  DB  : :  4:3  (10.6) ;  then 
the  construction  will  be  the 
same  as  in  the  last  example. 

The  calculation  will  like- 
G  wise  be  similar  to  the  last. 

The  results  are,  AF  =  36,235 ;  EF  =  9.378 ;  BDC  =  79° 
27';  CAB  =  41°  57';  ABC  =  63°  3' ;  AC  =  64.596;  BC  = 
48.447;  DC  =  43.927. 

4.  Given,  the  base  AB  500 ;  the  difference  of  the  sides  100 ; 
and  the  vertical  angle  ACB  72°,  to  determine  the  rest. 


O 


Construction.  On  the  base  AB 
describe  the  segment  of  a  circle 
containing  an  angle  equal  90°  +  ^ 
the  vertical  angle  (33.3) ;  place  in 
this  circle  the  right  line  AD  = 
the  difference  of  the  sides ;  pro- 


68 


FLANE  TRIGONOMETRY. 


duce  AD ;  join  DB ;  and  make  the  angle  DBC  =  BDC ;  then 
is  ABC  the  triangle  proposed. 

Draw  CE  at  right  angles  to  DB ;  then,  CD  being  =  CB 
(because  the  angle  CBD  =  CDE),  the  angle  DCB  is  evidently 
bisected  ;  and  the  angle 

ADB  =  DEC  +  DCE  =  90°  +  JACB ; 
also,  AD  =  AC  —  BC :  hence  the  construction  is  manifest. 

Calculation.  In  the  triangle  ABD,  AB,  AD,  and  the  an- 
gle ADB,  are  known;  whence  the  angle  BAD  may  be 
found ;  from  which  and  the  given  angle  ACB,  the  angle  ABC 
becomes  known.  Then  AB  and  all  the  angles  of  the  triangle 
being  known,  AC  and  BC  are  determined.  The  results  are 
BAC  =  44°  41',  ABC  =  63°  19',  AC  =  469.74;  and  BC 
=  369.74. 


5.  Given,  the  base  AB  465, 
the  vertical  angle  ACB  75° ; 
and  the  sum  of  the  sides  760, 
to  determine  the  rest. 

Construction.  On  AB  de- 
scribe a  segment  of  a  circle 
containing  half  the  vertical 
angle ;  from  A,  place  the  line 
AD  =  the  sum  of  the  sides, 
in    this    segment ;   join   DB ; 

and  make  the  angle  DBC  =  BDC.     Then  will  ACB  be  the 

triangle  proposed. 

Because  the  angle  DBC  m  BDC  ;  DC  =  BC ;  and  the  exte 
rior  angle  ACB  =  twice  ADB ;  also  AC  +  BC  =  AD. 

Calculation.  With  the  sides  AB,  AD,  and  angle  ADB ;  the 
angles  ABD  and  BAD  may  be  found ;  whence  ABC  becomes 
known.     Then  AC  and  BC  are  determined.     Results :  ABC 


SECTION   II.  69 

=  46°  45',  or  58°  15' ;  AC  =  350.64,  or  409.36 ;  BC  =  409.36 
or  350.64 

6.  Given,  the  base  AB  50 ;  the  line  DC,  drawn  from  the 
middle  of  the  base  to  the  vertex  40 ;  and  the  ratio  of  the 
sides,  AC  :  BC  : :  3  :  %  to  determine  the  sides 


Construction.  Divide  AB  in  E,  so  that  AE  :  EB  in  the  pro- 
posed ratio  of  AC  :  BC ;  produce  AB  to  F,  so  that  BF  shall 
be  a  third  proportional  to  AE  — -  EB  and  EB ;  from  the  cen- 
tre F,  through  E,  describe  the  arc  EC ;  from  the  centre  D, 
with  the  given  distance  DC,  describe  an  arc,  cutting  the 
former  in  C  ;  join  AC,  DC  and  BC  ;  then  ABC  is  the  triangle 
proposed. 

From  the  proportion  AE  —  EB  :  EB  : :  EB  :  BF,  we 

have  (18.5),  AE  :  EB  : :  EF  :  BF; 

therefore  (12.5)        AE  :  EB  : :  AF  :  EF ; 

consequently  (19.5), 

AF  :  EF  ::  EF  :  BF; 

whence  (F.  6),  AC  :  CB  : :  AE  :  EB. 

Calculation.  In  the  triangle  CDF,  we  have  all  the  sides  to 
find  ^  the  angle  FDC,  which  is  the  -J  sum  of  DAC  and  DCA  ; 
then,  with  that  J  sum  and  the  sides  AD,  DC,  the  angle  DAC 
and  side  AC  may  be  found. 

Result:  AC  =  55.50:  BC  =  37.00. 


70 


PLANE  TRIGONOMETRY. 


7.  Given*  the  sides  of  the 
triangle  ABC,  viz.,  AB  90,  AC 
80,  BC  70,  to  determine  the  dis- 
tances AD,  CD  and  BD,  to  a 
point  D,  which  is  so  situated 
that  the  angles  ADB  and  ADC 
are  70°  and  40°  respectively. 

Construction.  On  AB,  and 
on  the  side  opposite  to  C,  de- 
scribe the  segment  of  a  circle 
containing  an  angle  of  70° ; 
complete  the  circle;  at  the 
point  B  make  the  angle  ABE 
=  40°;  from  C,  through  E, 
where  BE  cuts  the  circle,  draw  CE,  and  produce  it  till  it 
cuts  the  circle  again  in  D,  the  point  required ;  join  DA,  DB, 
and  the  work  is  done. 

The  angle  ADE  ==  ABE  (21.3)  =  406  \  whence  the  con- 
struction is  manifest. 

Calculation.  Join  AE ;  then  the  angles  ABE,  BAE,  and 
the  side  AB>  are  known ;  whence  AE  may  be  found.  The 
angle  CAB  may  also  be  determined  from  the  three  sides; 
hence  CA,  AE,  and  the  contained  angle,  become  known ; 
from  which  ACE  and  AEC,  and  consequently  AED,  may  be 
found.  But  AED  =  ABD  (21.3) ;  therefore  the  angles  of  the 
triangle  ABD  become  known,  as  well  as  those  of  ADC  ;  from 
which  and  the  given  sides,  AD,  BD  and  CD  may  be  found. 

Result:    AD  =  82.915;    DB  =  73.406;  DC  =  123.178; 

and  DAB  =  50°  2'  6". 


8.  Given,  the  base  AB  50 ;  radius  of  the  circumscribing 
circle  30 ;  and  ratio  of  the  sides,  AC  :  BC  : :  3  :  2,  to  find 
the  sides. 


II 

'      I 

S  C 

A^ 

P     /        \ 

\ 

33 

1 

(; 


SECTION  II.  71 

Construction.  Divide  AB  in 
D  so  that  AD  :  DB  ::  3:2; 
bisect  AB  in  E ;  draw  EF  at 
light  angles  to  AB ;  from  the 
centre  A,  with  the  radius  of 
the  circumscribing  circle,  cut 
EF  in  F  ;  from  the  centre  F, 
with  the  radius  FA,  describe 
a  circle  cutting  FE  produced 
in  G;  join  GD,  and  produce 
it  to  meet  the  circle  in  C ;  join 
AC,  BC;  then  ABC  will  be  the  triangle  required.  The 
circle  will  pass  through  B  (1.3) ;  and  the  arc  AGB  is  bisected 
in  G ;  consequently,  the  angle  ACB  is  bisected  by  the  line 
CD  (27.3) ;  wherefore  (3.6), 

AC  :  BC  : :  AD  :  DB. 

Calculation.  In  the  right  angled  triangle  AEF,  AE  and 
AF  are  given,  from  which  EF  and  the  angle  AFE  are  deter- 
mined ;  but  AFE  =  2ACD  =  ACB.  In  the  triangle  EDG  ; 
EG  and  ED  being  known,  the  angle  EDG  or  BDC  is  deter- 
mined ;  whence  the  angles  of  the  triangle  ABC  are  known, 
and  the  sides  AC,  BC  determined. 

Result:  AC  =  59.447  ;  BC  *=  39.632. 


9.  Given,  one  angle  of  a  triangle  50° ;  the  sum  of  the  three 
sides  120;  and  the  radius  of  the  inscribed  circle  10,  to 
determine  the  sides  of  the  triangle. 

Construction.  Make  AB  =  \  the  sum  of  the  sides,  60 ;  the 
angle  BAN  =  the  given  angle  50° ;  bisect  that  angle  by  the 
line  AF,  meeting  BF,  which  is  drawn  at  right  angles  to  AB ; 
make  BC  =  the  given  radius  of  the  inscribed  circle,  10; 
through  C  draw  CD  parallel  to  BA,  meeting  AF  in  D ;  draw 
DE,  DG  at  right  angles  to  AB,  AN  respectively ;  from  the 
centre  D,  with  the  radius  DE,  describe  the  circle :  that  circle 


PLANE   TRIGONOMETRY. 

N 


will  touch  the  lines  AB,  AN,  in  E  and  G  (4.4).  On  the 
diameter  DF,  describe  the  semicircle  DHLF,  cutting  AB  in 
H  and  L ;  from  either  of  these  points  L  draw  LK,  touching 
the  circle  GEK  in  K,  and  cutting  AN  in  M ;  then  ALM  is 
the  triangle  proposed. 

Join  DL,  DK,  DM ;  LF,  MF ;  and  draw  FN,  FP  at  right 
angles  to  AN,  LK  respectively.  Now,  since  DE  =  DK,  the 
angles  at  E  and  K  are  right  ones,  and  DL  is  common  to  the 
triangles  DEL  and  DKL,  it  follows  that  the  angle  DLE  = 
DLK.  But  DLF  is  a  right  angle  (31.3) ;  hence,  DLE  +  FLB 
=DLF  (13.1)  ;  from  these  equals,  take  the  equals  DLE  and 
DLK ;  and  we  have  BLF  ==  KLF.  Thence,  the  angles  at 
B  and  P  being  right  ones,  it  is  obvious  that  LB  =  LP,  and 
FB  =  FP.  Again,  in  the  triangles  FAB  and  FAN,  we  have 
AF  common,  and  the  angles  of  the  one  respectively  equal  to 
those  of  the  other ;  hence  AN  ==  AB,  and  FN  =  FB  =  FP ; 
consequently  (47.1),  MN  =  MP. 

Now,  it  has  been  proved  that  LP  =  LB ;  consequently, 
ML  =  MN  +  LB; 
and,  therefore, 

AM  +  AL  +  ML  =  AB  +  AN=  2AB 


SECTION   IT.  73 

Calculation.  In  the  right  angled  triangles  AED,  ABF,  we 
have  the  angle  at  A,  and  the  lines  ED  and  AB  given;  from  which 
AE,  EB  and  BF  are  determined.  Then  the  angle  DCF  being 
a  right  one,  the  semicircle  on  DF  must  pass  through  C  (con- 
verse 31.3) ;  consequently,  CB.BF  =  LB.BH  (cor.  36.3).  If, 
now,  we  suppose  a  line  drawn  from  the  centre  of  the  semi- 
circle, cutting  EB  at  right  angles  in  I,  it  is  manifest  (2.6  and 
3.3)  that  EB  and  HL  are  both  bisected  in  I ;  whence  EL  = 
HB ;  and  EL.LB  =  CB.BF,  a  known  rectangle.  Hence  IL2 
{=  IB-  —  EL.LB  (5.2))  becomes  also  known. 

Result:  AL=50.306;  AM=31.139;  LM=38.555. 

10.  Given,  the  base  50;  difference  of  the  other  sides  10; 
and  radius  of  the  inscribed  circle  12 ;  to  determine  the  sides. 

Construction.  Make  AB= 
the  base  50,  and  bisect  it  in 
C;  lay  down  CD  =  J  the 
difference  of  the  sides  ;  at  D 
erect  a  perpendicular  DE  = 
radius  of  the  inscribed  circle 
A~~  c    D  B      12;  from  E,  with  the  dis- 

tance ED,  describe  a  circle ;  through  A  and  B  draw  the  lines 
AH,  BH,  touching  the  circle  in  F  and  G ;  ABH  is  the  triangle 
required. 

Join  EF,  EG;  then  (47.1)  AD  =  AF  ;  BD  =  BG;  FH  = 
GH;  consequently, 

AH  —  BH  =  AD  —  BD  -  2CD. 

Calculation.  With  AD,  DE,  and  BD,  DE,  find  the  angles 
BAE,  ABE,  from  which  BAH  and  ABH  are  known ;  and 
thence  the  sides  AH  and  BH. 

Result:  AH  ==  45.79;  BH  =  35.79. 

11.  Given,  the  perimeter  of  a  triangle  120;  radius  of  the 
inscribed  circle  10 ;  and  vertical  angle  70°,  to  determine  the 
sides. 

10  G 


74 


PLANE    TRIGONOMETRY. 


M 


l\/?& 

^2B^>^ 

A             H 

E                    /J) 

K 


Construction.  Make  AB= J  the  perimeter  60 ;  at  B  erect  the 
perpendicular  BC=radius  of  the  inscribed  circle  10;  through 
C,  draw  CG  parallel  to  AB ;  make  the  angle  BCD  =  comple- 
ment of  half  the  vertical  angle,  55° ;  bisect  AD  in  E ;  draw 
EF  at  right  angles  to  AD,  meeting  CD  produced  in  F ;  from 
F,  as  a  centre,  through  A  or  D,  describe  the  arc  AGB,  cut- 
ting the  line  GC ;  from  one  of  the  intersections  G,  draw  GH 
at  right  angles  to  AD;  from  G,  with  the  radius  GH, describe 
the  circle  HLM ;  through  A  and  D  draw  AI,  DI,  touching 
the  circle  in  L  and  M ;  then  ADI  is  the  triangle  required. 

Join  AG,  DG  and  IG;  then  it  is  obvious  that  those  lines 
bisect  the  angles  DAI,  ADI  and  AID  (26.1);  consequently, 

DAG  +  ADG  +  AIG  =  a  right  angle  (32.1) ; 

that  is,  AIG  is  the  complement  of  DAG  +  ADG.  Now,  since 
EFD  =  the  angle  at  the  circumference,  which  stands  on 
AGD  ;  it  follows  that 

EFD  +  AGD  =  2  right  angles  (22.3)  =  GAD  +  GDA  +  AGD 

(32.1);  wherefore. 

EFD  =  GAD  +  GDA; 

their  complements  are  therefore  equal;  that  is,  EDF  or  CDB 
—  LIG.     Consequently,  the  whole  angle  AID  is  twice  the 


SECTION   II.  79 

complement  of  DCB.  Also,  the  angles  LIG  and  ILG,  being 
respectively  equal  to  BDC  and  DBC,  and  LG  =  BC,  the  side 
IL  =  DB  (26.1);  hence  the  semiperimeter  of  the  triangle 
ADI  =  AB. 

Calculation.  Draw  FK  parallel  to  AB,  meeting  GH  pro- 
duced in  K,  and  join  FG ;  then,  in  the  triangle  DBC,  having 
BC  and  all  the  angles,  BD  is  found ;  whence  ED  becomes 
known ;  from  which,  and  the  angles,  DF  and  EF  are  found ; 
then,  in  the  right  angled  triangle  FGIv,  FG  and  GK  are 
known ;  whence  FK  or  EH  becomes  known ;  whence  AH 
and  BH  are  known.* 

Result:  AD=45.719 ;  AI=27.02;  DI=47.261. 


12.  Given,  the  sides  of  the  triangle  ABC>  viz. :  AB  4G4, 
AC  418,  and  BC  385 ;  it  is  required  to  find  a  point  D  within 
the  triangle,  such  that  AD,  BD  and  CD  shall  be  to  each  other 
in  the  ratio  of  7,  6  and  5  respectively. 


Construction.  Divide  AB  in  the  point  E,  and  AC  in  G,  so 
that  AE : EB  : : 7  :  6 ;  and  AG  :  GC  : :  7  :  5  ;  produce  AB 
and  AC  to  F  and  H,  so  that  BF  and  CH  shall  be  third  pro- 
portionals to  AE  —  EB  and  EB,  and  to  AG  —  GC  and  GC 
respectively ;  from  the  centres  F  and  H,  with  the  distance? 
FE  and  HG,  describe  arcs  cutting  each  other  in  D ;  join  AD 


*  Examples  9  and  11  are  essentially  of  the  same  nature,  and  might  have 
Seen  solved  by  the  same  method ;  the  two  solutions  furnish  a  little  variety. 


f70  PLANE   TRIGONOMETRY. 

BD  and  CD ;  and  the  figure  is  constructed.     For,  as  was 
proved  in  the  6th  example, 

AD  :  BD  : :  AE  :  EB  : :  7  :  6 ; 

and  AD  :  CD  ::  AG  :  GC  : :  7  :  5; 

whence,  BD  :  CD  : :  6  :  5. 


Calculation.  Join  DF,  DH  and  FH ;  then,  in  the  triangle 
ABC,  we  have  all  the  sides  to  find  the  angle  BAC ;  then,  in 
the  triangle  AFH,  we  have  the  sides  AF,  AH,  and  the  in- 
cluded angle,  to  find  the  angle  AFH  and  side  FH ;  in  the 
triangle  FDH,  the  three  sides  are  then  known  to  find  the 
angle  DFH ;  whence  the  angle  AFD  becomes  known :  then, 
in  the  triangle  AFD,  we  have  the  sides  AF,  FD,  and  the 
contained  angle,  to  find  the  angle  FAD  and  the  side  AD  ;  from 
which  BD  and  CD  are  found  from  the  given  ratios. 

Results:  BAD  =  25°  59'  8";  CAD  =  25°  27'  15";  AD  = 
283.688 ;  BD  =  243.161 ;  CD  =  202.635. 

[The  following  ingenious  construction  of  this  problem,  which 
admits  of  a  simpler  calculation  than  that  already  given,  has 
been  kindly  furnished  the  author  by  Samuel  Alsop,  Principal 
of  Friends'  Select  School,  Philadelphia.] 

Construction.  Make  FE  =t 
6;  and  on  it  describe  FEG 
similar  to  CBA,  the  given 
triangle,  making 

FE:EG:FG::BC:BA:AC. 


s  On  FG  describe  the  triangle 
FGA,  making  FA  =  7,  and 
AG  a  fourth  proportional  to 
BC,  AB  and  5.  Upon  AE,  on 
the  same  or  any  other  scale,  lay  down  AB  =  464,  and  com- 
ulete  the  triangle  ABC.  Draw  BD  parallel  to  EF,  cutting 
VF  in  D,  which  will  be  the  point  required.     For,  join  CD; 


SECTION   II.  77 

draw  EH  parallel  to  BC ;  arid  join  FH.     Then,  since  AEH 
is  similar  to  FEG,  being  both  similar  to  BAC ; 

AE  :  EG  ::  EH  :  EP ; 

therefore  (6.6),  AEG  and  HEF  are  similar ;  and 

AB  :  BC  : :  AE  :  EH  : :  AG  :  FH. 

But  AB  :  BC  : :  AG  :  5 ; 

therefore,  FH  ==  5.     Consequently, 

AD  :  BD  :  CD  : :  AF  :  FE  :  FH  : :  7  :  6  :  5. 

Calculation.  In  the  triangle  ABC,  with  the  given  sides,  find 
the  angle  BAC  =  FGE  ;  also  find  AG,  GF  and  GE.  From 
the  three  sides  of  the  triangle  AGF,  find  the  angle  AGF; 
whence  AGE  becomes  known.  In  the  triangle  AGE,  find 
AE;  then 

AE  :  EF  : :  AB  :  BD ; 

from  which  AD  and  CD  are  found  from  the  given  ratios. 

13.  In  a  right-angled  isosceles  triangle,  the  hypothenuse  is 
30  yards  longer  than  one  of  the  sides ;  what  are  the  sides  ? 

Ans. :  hypoth.  102.4264  :  sides  72.4264. 

14.  The  hypothenuse  of  a  right-angled  triangle  is  75,  and 
the  sum  of  the  sides  is  105;  what  are  the  sides? 

Ans. :  60  and  45. 

15.  The  sides  of  a  triangle  are  in  the  ratio  of  4,  6,  and  7 ; 
and  the  line  bisecting  the  greatest  angle  is  20 ;  required  the 
sides.  Result :  22.87,  34.31,  40.02. 

16.  Given  the  perimeter  of  a  right-angled  triangle  120, 
and  the  radius  of  the  inscribed  circle  10 ;  required  the  sides 
of  the  triangle?  Ans. :  50,  40,  and  30. 

a* 


78*  PLANE  TRIGONOMETRY. 

17.  From  a  position  in  a  horizontal  plane,  I  observe  the 
angle  of  elevation  of  a  tower,  which  is  100  feet  high,  to  be 
60° ;  how  far  must  I  measure  back,  to  obtain  a  position  from 
which  the  elevation  shall  be  30°?  Ans.  115.47  feet. 

18.  A  person  on  the  top  of  a  tower  which  is  50  feet  in 
height,  observes  the  angles  of  depression  of  two  objects  on 
the  horizontal  plane,  which  are  in  the  same  straight  line  with 
the  bottom  of  the  tower,  to  be  30°  and  45°.  Determine  their 
distance  from  each  other  and  from  the  observer. 

Ans.  Distance  from  each  other  36.60  feet. 
From  the  observer  70.71,  and  100  feet. 

19.  From  the  top  of  a  tower,  whose  height  is  108  feet,  the 
angles  of  depression  of  the  top  and  bottom  of  a  vertical 
column,  standing  in  the  horizontal  plane,  are  found  to  be  30° 
and  60°  respectively.     Required,  the  height  of  the  column. 

Ans.  72  feet. 

20.  Suppose  the  angle  of  elevation  of  the  top  of  a  steeple 
to  be  40°,  when  the  observer's  eye  is  level  with  the  bottom, 
and  that  from  a  window  18  feet  directly  above  the  first  sta- 
tion the  angle  of  elevation  is  found  to  be  37°  30'.  Required, 
the  height  and  distance  of  the  steeple. 

Ans.  Height,  210.44  feet. 
Distance,  250.79  feet. 

21.  Two  columns,  80  and  100  feet  in  height,  standing  on 
a  horizontal  plane,  are  distant  from  each  other  220  feet ;  it  is 
required  to  find  a  point  in  the  line  joining  their  bases,  from 
which  the  angles  of  elevation  of  the  two  columns  shall  be 
equal.      Ans.  The  point  is  122|  feet  from  the  higher  column. 

22.  The  altitude  of  a  cloud  was  observed  to  be  34°  20',  and 
that  of  the  sun  in  the  same  direction  50° ;  also  the  distance 
of  the  shadow  of  the  cloud  from  the  station  of  the  observer 
measured  375  yards.     Determine  the  height  of  the  cloud. 

Ans.  600  yards. 


SECTION  II.  79* 

23.  In  a  plane  triangle  there  are  given,  the  base  60,  an 
adjacent  angle  55°  30',  and  the  ratio  of  the  side  opposite  the 
given  angle  to  the  other  unknown  side  6  to  5 ;  to  determine 
these  sides.  Ans.  50.047,  and  41.706. 

24.  From  a  station  in  a  horizontal  plane,  I  observed  the 
angle  of  altitude  of  the  summit  of.  a  cliff  which  bore  exactly 
north  to  be  47°  30'.  I  then  measured  N.  87  W.  283  feet,  and 
again  taking  the  angle  of  altitude,  found  it  to  be  40°  12'. 
What  was  the  height  of  the  cliff?  Ans.  354.53  feet. 

Remark.  In  the  solution  of  the  preceding  problem,  it  will 
assist  the  pupil  if  he  will  observe,  that  when  two  right  angled 
triangles  have  the  same  perpendicular,  their  bases  are  to  each 
other  as  the  cotangents  of  the  angles  at  the  base. 

25.  Three  ships  sailed  from  the  same  place  to  different 
ports  in  the  same  parallel  of  latitude;  the  first  sailed  directly 
south  55  leagues,  when  she  arrived  at  the  desired  port ;  the 
other  two  sailed  upon  different  courses,  between  the  south 
and  west,  till  they  arrived  at  their  destined  ports,  which  were 
57  leagues  asunder,  and  the  angle  included  by  their  courses 
at  the  port  sailed  from  was  38°.  Required,  the  course  and 
distance  run  by  each  of  the  two  latter  vessels. 

Ans.    S.  52°  12',  W.  89.75  leagues; 
and  S.  14°  12',  W.  56.73  leagues. 

26.  Walking  on  shore,  I  was  surprised  by  the  flash  of  a 
gun,  at  sea,  bearing  S.  56°  15'  E. ;  seven  seconds  after  the 
flash  I  heard  the  report,  and  four  seconds  after  that  I  heard 
the  echo  from  a  castle  bearing  from  me  S.  56°  15'  W.  Re- 
quired, the  distance  of  the  gun  and  castle;  sound  being  esti- 
mated to  pass  over  1142  feet  in  one  second  of  time. 

Ans.  Distance  of  gun,  7994  feet;  of  castle,  3005.51  feet. 

27.  In  a  right  angled  triangle  there  are  given,  one  of  the 
legs  94,  and  the  segment  of  the  hypothenuse  adjacent  to  the 
other  leg,  made  by  a  perpendicular  from  the  right  angle,  66, 
to  determine  the  triangle. 

Ans.  The  other  leg  is  93.56,  and  the  hypothenuse,  132.62. 


80*  PLANE  TRIGONOMETRY. 

28.  Having  given  two  sides  of  a  triangle,  40  and  50,  and 
the  line  drawn  from  the  included  angle  to  the  middle  of  the 
third  side,  34;  to  determine  the  third  side.  Ans.  59.80. 

Construction. —  Form  the  triangle  BAE,  making  AB  40, 
AE  50,  and  BE  68 ;  complete  the  parallelogram  ABCE,  draw 
the  diagonal  AC,  and  ABC  will  be  the  required  triangle. 

29.  At  three  points  in  the  same  horizontal  straight  line,  the 
angles  of  elevation  of  an  object  were  found  to  be  36°  50',  21° 
24',  and  14°,  the  middle  station  being  84  feet  from  each  of  the 
others.     Required,  the  height  of  the  object. 

Ans.  53.96  feet. 

30.  Tn  a  level  garden  there  are  two  lofty  firs,  having  their 
tops  ornamented  with  gilt  balls :  one  is  100  feet  high,  the  other 
80,  and  they  are  120  feet  distant  at  the  bottom.  Now,  the 
owner  wants  to  place  a  fountain  in  a  right  line  between  the 
trees,  to  be  equally  distant  from  the  top  of  each,  and  to  make 
a  walk  or  path  from  the  fountain,  in  every  point  of  which  he 
shall  be  equally  distant  from  each  of  the  balfe ;  also,  at  the 
end  of  the  walk  he  would  fix  a  pleasure-house,  which  should 
be  at  the  same  distance  from  each  ball,  as  the  two  balls  are 
from  each  other.     How  must  this  be  done? 

Ans.  From  bottom  of  taller  tree  to  fountain,  45  feet. 
From  ball  to  ball,  121.655  " 

Length  of  the  walk,  52.678   " 

From  bottom  of  taller  tree  to  house,  69.282   " 

31.  Three  objects,  A,  B,  and  C,  are  situated  in  the  same 
straight  line,  and  are  distant  from  D,  312,  150,  and  123  yards ; 
also,  the  distance  of  A  from  B  is  to  the  distance  of  C  from 
B  as  22  to  13.     How  far  is  B  from  A  and  C  ? 

Ans.  From  A  198  yards,  from  C  117  yards 

Construction. — Make  AD  =  312,  and  divide  it  in  E  so  that 

AE  :  ED  :  :  22  :  13.    On  DE  form  the  triangle  DEB,  making 

DB  150,  and  EB  =  §§  DC.     Join  AB,  produce  it  till  it  meets 

DC  drawn  parallel  to  BE  in  C,  and  the  figure  is  constructed. 


SECTION  II  81* 

32.  Given,  the  angles  of  elevation  of  an  object  taken  at 
three  positions,  A,  B,  and  C,  in  the  same  horizontal  straight 
line,  17°  46',  33°  41',  and  39°  6',  respectively ;  also,  from  A  to 
B  is  264  feet,  and  from  B  to  C  156  feet.  Required,  the  height 
of  the  object.  Ans.  133.33  feet. 

33.  There  are  three  towns,  A,  B,  and  C,  whose  distances 
apart  are  as  follows:  from  A  to  B  6  miles;  from  A  to  C 
22  miles ;  and  from  B  to  C  20  miles.  A  messenger  is  des- 
patched from  B  to  A,  and  has  to  call  at  a  town  D  in  a  direct 
line  between  A  and  C.  Now,  in  travelling  from  B  to  D,  he 
walks  uniformly  at  the  rate  of  4  miles  an  hour,  and  from  D 
to  A  at  the  rate  of  3  miles  an  hour.  Supposing  him  to  per- 
form his  journey  in  three  hours,  it  is  required  to  determine 
the  position  of  the  town  D. 

Ans.  The  distance  of  D  from  A  is  4.72  miles. 

In  the  above  example,  we  have  J  AD  +  £  BD  =  3,  or 
AD  +  |BD  =  9.  On  AC  lay  off  AE  =  9,  and  join  BE; 
then  in  the  triangle  BDE  the  side  BE  and  the  angle  BED 
become  known,  and  ED  :  DB  :  :  3  :  4.  Hence  the  point  D  is 
readily  determined. 

34.  The  lengths  of  three  lines  drawn  from  a  given  point  to 
three  angles  of  a  square  are,  35,  46,  and  50  yards ;  to  deter- 
mine a  side  of  the  square.  Ans.  59.95  yards. 

35.  Wishing  to  ascertain  the  length  of  a  tree  which  leaned 
in  the  plane  of  the  meridian,  I  measured  from  the  foot  of  the 
tree  north  85  feet,  when  I  found  the  angle  of  elevation  of  the 
top  to  be  35°.  I  then  took  a  second  station  50  feet  east  of 
the  former,  at  which  the  elevation  was  30°.  Required,  the 
length  of  the  tree.  Ans.  52.44  feet. 

11 


SECTION   III. 


SPHERICAL  TRIGONOMETRY. 

Article  45.  The  business  of  Spherical  Trigonometry  is, 
to  investigate  the  properties  of  triangles  formed  on  the  sur- 
face of  a  sphere,  by  the  arcs  of  circles  whose  planes  pass 
through  the  centre. 

As  the  diameter  of  a  circle  is  the  greatest  straight  line  in 
it  (15.3),  so  the  diameter  of  a  sphere  is  necessarily  the  great- 
est straight  line  in  it.  Hence,  when  a  plane  passes  through 
the  centre  of  the  sphere,  the  diameter  of  the  circle  which  is 
formed  by  the  section  of  this  plane  and  the  sperical  surface, 
is  greater  than  any  other  line  in  the  sphere  which  is  not  a 
diameter. 

A  plane  cutting  the  sphere,  but  not  passing  through  its 
centre,  forms,  by  its  section  with  the  spherical  surface,  a 
circle  whose  diameter  is  less  than  the  diameter  of  the  sphere. 
That  the  section  is  a  circle,  is  readily  inferred  from  14.3 ; 
and  that  the  diameter  of  that  circle  is  less  than  the  diameter 
of  the  sphere,  is  plain  from  15.3. 

Definition  1.  Those  circles  whose  planes  pass  through  the 
centre  of  the  sphere,  are  called  great  circles ;  but  circles 
whose  planes  do  not  pass  through  the  centre  of  the  sphere, 
are  called  less  circles. 

Corollary  1.  The  diameter  of  every  great  circle  is  also  a 
diameter  of  the  sphere. 

Cor.  2.  The  common  section  of  the  planes  of  two  great 
circles,  is  a  diameter  to  each  of  those  circles. 

(78) 


SECTION    III. 


79 


Cor.  3.  Every  great  circle  in  the  sphere  divides  every 
other  great  circle  into  two  equal  parts. 

Def.  2.  The  axis  of  a  circle  is  the  right  line  which  passes 
through  its  centre,  and  is  at  right  angles  to  the  plane  of  the 
circle ;  and  the  poles  of  a  circle  are  the  points  where  its  axis 
meets  the  surface  of  the  sphere. 

Def.  3.  A  spherical  angle,  or  the  angle  formed  by  two 
great  circles,  is  the  inclination  of  their  planes. 

Cor.  When  two  great  circles  are  at  right  angles  to  each 
other,  each  of  them  passes  through  the  poles  of  the  other ; 
and  if  they  pass  through  the  poles  of  each  other,  they  are  at 
right  angles.  Also,  when  the  plane  of  a  great  circle  is  at 
right  angles  to  the  plane  of  a  less  one,  the  former  circle 
passes  through  the  poles  of  the  latter.  For  the  axis  of  every 
circle  passes  through  the  centre  of  the  sphere,  and  is  at  right 
angles  to  the  plane  of  its  own  circle. 

Def.  4.  A  spherical  triangle  is  formed  by  the  arcs  of  three 
great  circles,  each  of  which  cuts  the  other  two,  but  in  such 
manner  that  each  of  the  arcs  composing  the  triangle  is  less 
than  a  semicircle. 

Def.  5>  If  AD  and 
DF,  two  quadrants  of 
great  circles,  are  placed 
at  right  angles  to  each 
other ;  and  through  the 
points  A,  F,  two  other 
great  circles,  AE,  FB; 
are  described,  cutting 
each  other  in  C;  the 
triangles  ABC,  FEC 
are  called  complementat 
triangles. 

Art.  46.  The  arc  of  a  great  circle,  intercepted  betweer 
another  great  circle  and  its  pole,  is  a  quadrant. 


80 


SPHERICAL  TRIGONOMETRY. 
D 


Let  AEBF  be  a  great  circle,  whose  centre  is  C,  and  axis 
DCG,  its  poles  being  D  and  G ;  DAGB  another  great  circle 
passing  through  the  axis  DCG;  these  great  circles  are  at 
right  angles  to  each  other  (17.2  sup.),  and  CA  their  common 
section  at  right  angles  to  CD ;  hence  the  arcs  AD,  BD,  AG 
and  BG,  are  quadrants. 

Art.  47.  The  angle  made  by  two  great  circles  is  measured 
by  the  arc  intercepted  between  them,  at  the  distance  of  90° 
from  the  angular  point. 

Let  ACB,  ADB,  be 
two  semicircles,  whose 
common  section  passes 
through  E,  the  centre 
of  the  sphere ;  from  E, 
draw  EC,  ED,  at  right 
angles  to  AB,  one  in  the 
plane  ACB,  the  other  in  the  plane  ADB ;  and  let  the  plane 


SECTION   III. 


81 


passing  through  EC,  ED,  cut  the  surface  of  the  sphere  in  the 
arc  CD ;  CD  is  part  of  a  great  circle  (Def.  1),  and  AE  is  at 
right  angles  to  its  plane  (4.2  sup.) ;  consequently,  AC  and  AD 
are  quadrants;  and  A,  B  are  the  poles  of  CD.  Also,  the 
inclination  of  the  planes  ACB,  ADB  is  the  angle  CED  (Def. 
4.2  sup.) ;  and  that  angle  is  measured  by  the  arc  CD. 

Cor.  Since  the  plane  of  CED  is  at  right  angles  to  AB,  ana 
consequently  to  each  of  the  planes  ADB,  ACB  (17.2  sup.),  it 
must  pass  through  the  axes  of  those  planes ;  and  therefore 
the  circle  DC,  continued,  must  pass  through  the  poles  of  ADB 
and  ACB.  Those  poles  being  90  degrees  from  their  respect- 
ive circles,  the  arc  intercepted  between  them  is  manifestly 
equal  to  CD,  the  measure  of  the  spherical  angle  CAD. 


Art.  48.  In  the  complemental  triangles  ABC,  FCE ;  AC 
is  the  complement  of  CE  ;  BC  is  the  complement  of  FC  ;  AB 
of  the  angle  at  F ;  and  the  angle  at  A  of  the  side  FE. 

For,  since  FD  and 
AD  are  quadrants  at 
right  angles  to  each 
other,  F  is  the  pole  of 
AD,  and  A  is  the  pole 
ofFD  (Art.  46);  hence 
FB,  AE  are  also  quad- 
rants; consequently,  BD 
is  the  measure  of  the 
angle  at  F,  and  DE  of 
the  angle  at  A  ;  whence 
the  proposition  is  ob- 
vious. 

Art.  49.  In  isosceles  spherical  triangles,  the  angles  oppo- 
site the  equal  sides  are  equal. 

Let  ABC  be  a  spherical  triangle,  whose  sides  AB  and  AC 
are  equal ;  it  is  to  be  proved  that  the  angles  ABC  and  ACE 
are  also  equal. 


82  SPHERICAL  TRIGONOMETRY. 

Take  D  the  centre  of  the  sphere, 
and  join  DA,  DB  and  DC ;  and 
in  the  plane  of  ADB,  draw  AE 
at  right  angles  to  DB.  In  like 
manner,  in  the  plane  ADC,  draw 
AF  at  right  angles  to  DC.  Then, 
since  the  arc  AB  is  equal  to  AC, 
"j!j  the  angle  ADB  at  the  centre  of 

the  sphere  is  equal  to  the  angle  ADC ;  therefore  the  triangles 
ADE,  ADF,  right  angled  at  E  and  F,  having  two  angles  of 
the  one  respectively  equal  to  two  angles  of  the  other,  and 
the  side  AD,  opposite  the  right  angle  in  each,  common  to 
both ;  have  the  sides  AE,  AF,  adjacent  to  the  right  angles, 
also  equal  (26.1). 

Again,  in  the  plane  BDC,  draw  EG  and  FG  at  right  angles 
to  DB  and  DC  respectively,  and  let  them  meet  in  G.  Then, 
because  AE  and  EG  are  both  at  right  angles  to  DB,  the  line 
DB  is  at  right  angles  to  the  plane  which  passes  through  AE 
and  EG  (4.2  sup.) ;  and  therefore  the  plane  DBC  is  at  right 
angles  to  the  plane  AEG  (17.2  sup.).  In  like  manner,  the 
plane  DBC  is  proved  to  be  at  right  angles  to  the  plane  AFG; 
consequently,  the  line  AG,  the  common  section  of  the  planes 
AEG,  AFG,  is  at  right  angles  to  the  plane  DBC  (18.2  sup.) ; 
wherefore  the  angles  AGE,  AGF  are  right  angles  (1  Def.  2 
sup.) 

Now,  the  right  angled  triangles  AGE,  AGF,  having  the 
perpendicular  AG  common,  and  the  hypothenuse  AE  equal 
the  hypothenuse  AF,  must  have  their  bases  EG,  FG,  also 
equal  (47.1) ;  and  therefore  the  angles  AEG,  AFG,  likewise 
equal  (8.1) ;  that  is,  the  spherical  angles  ABC,  ACB,  are 
equal.  Q.  E.  D. 

Art.  50.  If  two  angles  of  a  spherical  triangle  are  equal, 
the  sides  opposite  to  them  are  also  equal. 

Let  the  spherical  angles  ABC,  ACB  be  equal ;  then  the 
sides  AB,  AC  shall  be  also  equal.  ^ 


SECTION  III.  83 

Making  the  same  construction  as  in  the  last  article,  w 
have,  as  before,  the  angles  AGE,  AGF,  both  right  angles  s 
also  the  angles  AEG  and  AFG,  which  are  the  same  as  the 
spherical  angles  ABC  and  ACB,  likewise  equal,  and  the  side 
AG  common  to  the  triangles  AGE,  AGF ;  therefore  AE  is 
equal  to  AF  (26.1).  Then,  in  the  right  angled  triangles 
ADE,  ADF,  we  have  the  perpendiculars  AE,  AF  equal,  and 
the  hypothenuse  AD  common ;  wherefore  DE  is  equal  to  DF 
(47.1),  and  consequently  the  angle  ADE  equal  to  ADF  (8.1)  : 
whence  AB  is  equal  to  AC  (26.3). 

Q.  E.  D. 

Art.  51.  Any  two  sides  of  a  spherical  triangle  are  together 
greater  than  the  third. 

Let  ABC  be  a  spherical 
triangle;  any  two  of  its 
sides  taken  together  are 
greater  than  the  third. 

Take  D  the  centre  of  the 
sphere,  and  join  DA,  DB 
and  DC.  Then  the  solid 
angle  at  D  is  contained  by 
the  three  plane  angles 
ADB,  ADC  and  BDC,  of  which  any  two  taken  together  are 
greater  than  the  third  (20.2  sup.)  ;  therefore  any  two  of  the 
arcs  which  measure  those  angles  are  likewise  together  greater 
than  the  third.  Q.  E.  D. 


Art.  52.  The  three  sides  of  a  spherical  triangle  are  toge- 
ther less  than  the  circumference  of  a  circle. 


Let  ABC  be  a  spherical 
triangle;  the  sides  AB  + 
AC  +  BC  are  less  than 
360°.  Continue  two  of 
those  sides  AC,  AB,  till 


84  SPHERICAL   TRIGONOMETRY. 

they  meet  in  D ;  then  ACD  and  ABD  are  semicircles  (Art. 
45,  Cor.  3,  Def.  1).  But  BD  +  CD  are  greater  than  BC.  If 
to  these  unequal  quantities  we  add  AB  +  AC,  we  have  ABD 
+  ACD  greater  than  BC  +  AB  +  AC ;  that  is,  AB  +  AC  + 
BC  are  less  than  two  semicircles,  or  360°. 

Q.  E.  D. 

Art.  53.  In  any  spherical  triangle  having  unequal  angles, 
the  greater  angle  has  the  greater  side  opposite  to  it. 

In  the  spherical  triangle  ABC,  let  the  angle  ABC  be  greater 
than  ACB ;  and  take  CBD  =  BCA ;  then  (Art.  50)  BD=CD ; 
consequently,  AC  =  BD  +  AD ;  but  (Art.  50)  BD  +  AD  are 
greater  than  AB ;  that  is,  AC  is  greater  than  AB. 

Q.  E.  D. 

Conversely :  If  the  side 
AC  is  greater  than  AB,  the 
angle  ABC  is  greater  than 
ACB.  For  if  it  is  not 
greater,  it  is  equal  or  less. 
If  the  angles  were  equal, 
the  opposite  sides  would  also  be  equal  (Art.  49) ;  and  if  ACB 
was  greater  than  ABC,  the  side  AB  would  be  greater  than 
AC. 

Art.  54.  If  the  angular  points  of  a  spherical  triangle  are 
made  the  poles  of  three  great  circles,  these  three  circles,  by 
their  intersections,  will  form  a  triangle,  which  is  said  to  be 
supplemental  to  the  former ;  and  the  two  triangles  are  such, 
that  the  sides  of  the  one  are  the  supplements  of  the  arcs 
which  measure  the  angles  of  the  other. 

Let  A,  B,  C,  the  angular  points  of  the  triangle  ABC,  be 
the  poles  of  the  great  circles  FE,  DE,  DF,  which  form  the 
triangle  FED  ;  and  let  the  sides  of  the  former  triangle  be 
produced  till  they  meet  those  of  the  latter.  Now,  since  A 
and  C  are  the  poles  of  EF  and  DF  respectively,  the  distances 


SECTION    III. 


85 


from  F  to  A  and  from  F 
to  C  are  quadrants  (Art. 
46) ;  hence  F  is  the  pole 
of  K ACL.  In  the  same 
manner  it  is  proved  that 
D  and  E  are  the  poles  of 
NBCH  and  GABM ;  con- 
sequently, EM  +  LF  = 
180°;  thatis,EFandLM 
are  supplements  to  each 
other.  In  like  manner  it 
may  be  proved  that  DE 
and  GH;  DF  and  KN; 
AC  and  KL;  AB  ar.d 
GM;  BC  and  NH,  are 
respectively  supplements  to  each  other.  But  ML,  GH  and 
KN  are  the  measures  of  the  angles  A,  B  and  C  ;  also  KL, 
NH  and  GM  are  the  measures  of  the  angles  F,  D  and  E  (Art. 
47).     Hence  the  proposition  is  manifest?  Q.  E.  D. 

Cor.  Since  the  sides  FE,  FD  and  DE,  together  with  the 
measures  of  the  angles  A,  B  and  C,  are  equal  to  three  semi- 
circles, or  540° ;  and  the  three  sides  of  any  spherical  triangle 
are  together  less  than  two  semicircles,  or  360°  (Art.  52),  it 
follows  that  the  three  angles  of  the  triangle  ABC  are  more 
than  180°,  but  less  than  540°. 


Aet.  55.    Let  AEB  and 

AHB  be  semicircles,  whose 

planes  are  at  right  angles  to 

each    other ;    and   AB    the 

common    section    of    those 

planes,  a   diameter   to    the 

sphere;     AH,    HB,    quad- 

P  e  rants;     and    C    any   other 

point  than  H  in   the  semicircle  AHB;  then  CD,  CE,  CF, 

being  arcs  of  great  circles,  intercepted  between  the  point  C 

12 


86  SPHERICAL   TRIGONOMETRY. 

and  the  semicircle  AEB  ;  the  arc  CA,  which  passes  through 
H,  is  greater,  and  CB,  the  remaining  part  of  the  semicircle, 
is  less,  than  any  other  arc  contained  between  C  and  AEB ; 
also  those  nearer  to  CHA  are  greater  than  those  which  are 
more  remote. 

Draw  CG  and  HI  at  right  angles  to  AB ;  then  is  CG  at 
right  angles  to  the  plane  AEB  (Def.  2,  2  sup.) ;  hence,  GD, 
GE,  GF,  being  drawn,  the  angles  CGD,  CGE,  CGF,  are  all 
right  ones.  And  AH,  BH,  being  quadrants,  H  is  evidently 
the  pole,  and  I  the  centre,  of  AEB ;  consequently,  GA  is  the 
greatest,  and  GB  the  least,  of  all  the  straight  lines  drawn 
from  G  to  the  circumference ;  and  GD  is  greater  than  GE ; 
and  GE  than  GF  (7.3).     Now, 

AO-CGM-GA8; 
and  .  DC3  =  CG3  +  GD3; 

of  which  GA3  is  greater  than  GD3 ;  wherefore  AC2  is  greater 
than  DC3,  and  AC  greater  than  DC.  But  the  arcs  AC  and 
DC  are  each  less  than  a  semicircle ;  and,  therefore,  the  greater 
chord  subtends  the  greater  arc ;  that  is,  the  arc  AC  is  greater 
than  DC.  In  the  same  manner  it  may  be  proved  that  the  arc 
DC  is  greater  than  EC,  that  EC  is  greater  than  FC,  and 
FC  greater  than  BC.  Q.  E.  I). 

Art.  56.  In  a  right  angled  spherical  triangle,  the  sides 
containing  the  right  angle  are  of  the  same  affection  as  the 
angles  opposite  to  them.* 

Let  ACB  be  a  spherical  triangle,  right  angled  at  A ;  and 
let  AC,  AB  be  continued  till  they  meet  in  D  ;  and  bisect  ABD 
in  E ;  then  ACD,  ABD  are  semicircles  (Art.  45,  Cor.  3  to 
Def.  1) ;  and  AE  is  a  quadrant.  But  the  angle  at  A  being  a 
right  one,  AEB  passes  through  the  pole  of  AC  (Art.  45,  Cor. 


*  Sides  are  said  to  be  of  the  same  affection  when  they  are  both  less  or 
both  greater  than  quadrants ;  the  same  is  said  of  angles  when  they  are 
both  less  or  both  greater  than  right  angles.  A  side  and  an  angle  are  also 
of  the  same  affection  when  the  former  is  less  or  greater  than  a  quadrant, 
and  the  latter  less  or  greater  than  a  right  angle. 


SECTION   III.  87 

to  Def.  3).     Consequently  E  is  the  pole  of  AC  (Art.  46) ;  CE 
is  a  quadrant,  and  ACE  a  right  angle. 

Now,  AC  being  taken 
less  than  a  quadrant,  the 
angle  ACB  will  be  less  or 
greater  than  ACE,  ac- 
cording as  B  lies  be- 
tween A  and  E,  or  between 
E  and  D;  that  is,  when  AB 
is  less  than  a  quadrant,  the  angle  ACB  is  less  than  a  right 
angle ;  and  when  AB  is  more  than  a  quadrant,  ACB  is  more 
than  a  right  angle.  And  if  we  suppose  ACB  to  be  less  than 
a  right  angle,  it  is  manifest  that  AB  is  less  than  a  quadrant ; 
and,  if  greater,  greater. 

Again,  in  the  right  angled  triangle  DCB,  in  which  DC  is 
greater  than  a  quadrant,  it  is  manifest  that  the  angle  DCB  is 
greater  or  less  than  a  right  angle,  according  as  DB  is  greater 
or  less  than  a  quadrant ;  and  vice  versa.  Q.  E.  D. 

Art.  57.  When  the  sides  of  a  right  angled  spherical  tri- 
angle, about  the  right  angle,  are  of  the  same  affection,  the 
hypothenuse  is  less  than  a  quadrant ;  but  when  those  sides 
are  of  different  affections,  the  hypothenuse  is  more  than  a 
quadrant. 

Retaining  the  construction  used  in  the  last  article,  and 
bisecting  ACD  in  G,  we  have  G  the  pole  of  ABD,  and  CE  a 
quadrant  as  before.  But  CB  is  either  greater  or  less  than 
CE,  according  as  it  is  nearer  to  or  farther  from  CGD  than 
CE  is  (Art.  55) ;  that  is,  the  hypothenuse  is  less  than  a  quad- 
rant when  the  sides  are  both  less  or  both  greater  than  a 
quadrant ;  but  the  hypothenuse  is  greater  than  a  quadrant 
when  one  side  is  less,  and  the  other  greater,  than  a  quadrant. 

Q.  E.  D. 

Cor.  1.  Conversely,  when  the  hypothenuse  of  a  right  angled 
spherical  triangle  is  less  than  a  quadrant,  the  sides  are  of  the 
same  affection ;  but,  when  the  hypothenuse  is  greater  than  a 
quadrant,  the  sides  are  of  different  affections. 


bd 


SPHERICAL  TRIGONOMETRY. 


Cor.  2.  Since  the  oblique  angles  of  a  right  angled  spherical 
triangle  are  of  the  same  affection  as  the  opposite  sides  (Art. 
56) ;  therefore,  according  as  the  hypothenuse  is  greater  or 
less  than  a  quadrant,  the  oblique  angles  will  be  of  different, 
or  the  same  affection. 

Cor.  3.  Because  the  sides  are  of  the  same  affection  as  their 
opposite  angles ;  therefore,  when  an  angle  and  the  side  adja- 
cent are  of  the  same  affection,  the  hypothenuse  is  les-s  than  a 
quadrant,  and  vice  versa. 


Art.  58.  In  any  right  angled  spherical  triangle,  as  radius 
is  to  the  sine  of  an  oblique  angle,  so  is  the  sine  of  the  hy- 
pothenuse to  the  sine  of  the  opposite  side. 

Let  ABC  be  the  triangle, 
right  angled  at  B ;  take  D 
the  centre  of  the  sphere ; 
join  DA,  DB,  and  DC ;  in 
the  plane  ADC,  draw  CE 
at  right  angles  to  DA ; 
from  E,  draw  in  the  plane 
ADB,  the  line  EP  at  right 
angles  to  DA,  meeting  DB 
in  F ;  and  join  CF. 

Then,  the  lines  EC  and 
EF  being  both  at  right  angles  to  DA,  the  plane  CEF  is  at 
right  angles  to  DA  (4.2  sup.) ;  consequently,  the  planes  ADB 
and  CEF  are  at  right  angles  to  each  other  (17.2  sup.).  But 
the  plane  DBC  is,  by  hypothesis,  at  right  angles  to  DAB ; 
hence  the  planes  CEF  and  DBC,  being  both  at  right  angles 
to  DAB,  their  common  section  FC  is  also  at  right  angles  to 
the  same  plane  (18.2  sup.) ;  wherefore  DFC  and  EFC  are 
right  angles.  Hence  CF  is  the  sine  of  CB ;  also  CE  is  the 
sine  of  CA ;  and  the  angle  CEF  is  the  inclination  of  the 
planes  CDA  and  BDA ;  that  is,  Cl^F  =  the  spherical  angle 
CAB.     Now  (Art.  28), 

As  radius  :  sine  of  CEF  : :  CE  :  CF ; 


SECTION    III.  89 

that  i?, 

As  radius  :  sine  of  CAB  : :  sine  of  AC  :  sine  of  BC. 

Q.  E.  D. 

Art.  59.  In  any  oblique  angled  spherical  triangle,  the  sines 
of  the  angles  are  to  each  other  as  the  sines  of  the  opposite 
sides. 

Let  ABC  be  the  trian- 
gle ;  and  through  C  de- 
scribe the  arc  CD  of  a 
B  great  circle,  at  right  an- 
gles to  AB  ;  then,  by  last 
article, 

As  sin  of  A  :  radius  : :  sin  DC  :  sin  AC ;  ' 

radius  :  sin  B  : :  sin  BC  :  sin  DC ; 

therefore  (23.5), 

sin  A  :  sin  B  : :  sin  BC  :  sin  AC. 

Q.  E.  D. 


Art.  60.  In  any.  right 
angled  spherical  triangle, 
as  radius  is  to  the  sine  of 
one  of  the  sides,  so  is  the 
tangent  of  the  adjacent 
angle  to  the  tangent  of 
the  opposite  side. 

Let  ABC  be  the  trian- 
gle, right  angled  at  B; 
take  D  the  centre  of  the 
sphere,  and  join  DA,  DB, 
DC;  in  the  plane  ADB, 
draw  BE  at  right  angles 


90 


SPHERICAL   TRIGONOMETRY. 


to  DA ;  from  E,  draw  EF,  in  the  plane  ADC,  at  right  angles 
to  AD,  and  meeting  DC  produced  in  F  ;  and  join  FB.  Then 
EB  and  EF  being  at  right  angles  to  DE,  the  plane  FEB  is 
at  right  angles  to  DE  (4.2  sup.) ;  consequently,  the  plane 
ADB,  which  passes  through  DE,  is  at  right  angles  to  the 
plane  FEB  (17.2  sup.) ;  therefore  the  common  section  BF  of 
the  planes  EBF  and  DBC,  is  at  right  angles  to  the  plane 
DAB  (18.2  sup.) ;  whence  EBF  and  DBF  are  right  angles ; 
and,  consequently,  BF  is  the  tangent  of  BC ;  BE  is  also  the 
sine  of  AB  ;  and  the  angle  BEF  the  same  as  the  spherical 
angle  BAC.     Now  (Art.  28), 


that  is, 


As  radius  :  tan  BEF  : :  BE  :  BF ; 


As  radius  :  tan  BAC  : :  sin  AB  :  tan  BC ; 


and  alternately  (16.5), 

As  radius  :  sin  AB 


tan  BAC  :  tan  BC. 

Q.E.D. 


Art.  61.  If  two  right  angled  spherical  triangles  have  the 
same  perpendicular,  the  sines  of  the  bases  are  to  each  other 
^  reciprocally  as  the  tan- 

gents   of    the    adjacent 
angles. 

Let  ADC  and  BDC  be 
t      the   triangles ;    DC    the 
common   perpendicular; 
then  (Art.  60), 

As  sin  AD  :  radius  : :  tan  DC  :  tan  A  ; 

and  As  radius  :  sin  BD  : :  tan  B  :  tan  DC ; 

whence  (23.5), 


AD  :  sin  BD  : :  tan  B  :  tan  A. 


Q.  E.  D. 


SECTION   III. 


91 


Art.  62.  In  any  right  angled  spherical  triangle  ;  as  radius 
is  to  the  cosine  of  the  angle  at  the  base,  so  is  the  tangent  of 
the  hypothenuse  to  the  tangent  of  the  base. 

p  .      Let  ABC  be  the  tri- 

angle ;  B  the  right,  an- 
gle ;  and  FCE  the  com- 
plemental  triangle. 
Then  (Art.  60), 

As  radius  :  sin  FE  : : 
tan  F  :  tan  CE  ; 

that  is  (Art.  48), 

As    radius  :  cos    A  : : 
cotan  AB  :  cotan  AC. 

B  But  (Art.  23.4), 

As  tan  P  :  tan  Q  : :  cotan  Q  :  cotan  P ; 

consequently, 

As  radius  :  cos  A  : :  tan  AC  :  tan  AB. 

Q.  E.  D. 


Art.  63.  If  two  right  angled  spherical  triangles  have  the 
same  perpendicular,  the  cosines  of  the  vertical  angles  are  to 
each  other  reciprocally  as  the  tangents  of  the  hypothenuses. 

Let  ADC,  BDC  (see  fig.  on  opposite  page),  be  the  triangles 
right  angled  at  D ;  then  (Art.  62), 

As  cos  ACD  :  radius  : :  tan  DC  :  tan  AC ; 

and         As  radius  :  cos  BCD  : :  tan  BC  :  tan  DC ; 

therefore  (23.5), 

As  cos  ACD  :  cos  BCD  : :  tan  BC  :  tan  AC. 

Q.KD 


92 


SPHERICAL   TRIGONOMETRY. 


Art.  64.  In  any  right  angled  spherical  triangle ;  as  radius 
is  to  the  cosine  of  the  hypothenuse,  so  is  the  tangent  of  either 
angle  to  the  cotangent  of  the  remaining  angle. 

In  the  triangle  FCE  (see  fig.  on  opposite  page),  we  have 
(Art.  60), 

As  radius  :  sin  CE  : :  tan  FCE  :  tan  FE ; 
that  is  (Art.  48), 

As  radius  :  cos  AC  : :  tan  ACB  :  cot  BAC  : :  (Art.  23.4) 
tan  BAC  :  cot  ACB. 

Q.  E.  D. 

Art.  65.  In  any  right  angled  spherical  triangle ;  as  radius 
is  to  the  cosine  of  one  of  the  sides,  so  is  the  cosine  of  the 
other  side  to  the  cosine  of  the  hypothenuse. 

In  the  right  angled  triangle  FCE,  we  have  (Art.  57), 

As  radius  :  sin  F  : :  sin  FC  :  sin  CE ; 

that  is  (Art.  48), 

As  radius  :  cos  AB  : :  cos  BC  :  cos  AC. 

Q.  E.  D. 

Art.  66.  If  two  right  angled  spherical  triangles  have  the 
C  same  perpendicular,  the 

cosines  of  the  hypothe- 
nuses  are  to  each  other 
as    the    cosines   of    the 
*      bases. 

Taking  ADC  and  BDC 
as    the    triangles    right 
angled  at  D,  we  have  (Art.  65), 

As  radius  :  cos  DC  : :  cos  AD  :  cos  AC ; 

and  As  radius  :  cos  DC  : :  cos  BD  :  cos  BC ; 

whence  (16.5), 


SECTION   III. 
cos  AD  :  cos  BD  : :  cos  AC  :  cos  BC. 


Q.  E.  D. 


Art.  67.  In  any- 
right  angled  spherical 
triangle;  as  radius  is 
to  the  sine  of  either 
oblique  angle,  so  is  the 
cosine  of  the  adjacent 
side  to  the  cosine  of  the 
opposite  angle. 

In   the  right  angled 
triangle  FCE,  we  have 
b  (Art.  58), 

As  radius  :  sin  FCE  : :  sin  CF  :  sin  FE ; 

that  is  (Art.  48), 

As  radius  :  sin  ACB  : :  cos  BC  :  cos  BAC. 

Q.  E.  D. 

Art.  68.  In  two  right  angled  spherical  triangles,  A  CD, 
BCD  (fig.  p.  92),  having  the  same  perpendicular  CD,  the  co- 
sines of  the  angles  at  the  base  are  to  each  other  as  the  sines 
of  the  vertical  angles. 

By  Art.  67  and  A.  5, 

As  cos  DAC  :  cos  DC  : :  sin  ACD  :  radius ; 
and,  by  same  article, 

As  cos  DC  :  cos  DBC  : :  radius  :  sin  BCD ; 

consequently  (22.5), 

As  cos  DAC  :  cos  DBC  : :  sin  ACD  :  sin  BCD. 

Q.  E.  D. 
13 


94  SPHERICAL  TRIGONOMETRY. 

Art.  69.  The  same  things  being  supposed  as  in  the  last 
article,  the  tangents  of  the  bases  are  to  each  other  as  the 
tangents  of  the  vertical  angles. 

By  Art.  60, 

As  radius  :  sin  CD  : :  tan  ACD  :  tan  AD ; 

and  As  radius  :  sin  CD  : :  tan  BCD  :  tan  BD ; 

consequently  (11  and  16.5), 

tan  ACD  :  tan  BCD  : :  tan  AD  :  tan  BD. 

Q.  E.  D. 


Art.  70.  In  two 
right  angled  spheri- 
cal triangles  ABC, 
ADC,  having  the 
same  hypothenuse 
AC,  the  cosines  of  the 
bases  are  to  each 
other  reciprocally  as 
the  cosines  of  the  per- 
pendiculars. 
For  (Art.  65), 

As  radius  :  cos  AB  : :  cos  BC  :  cos  AC ; 
and  from  the  same  article  inverted  (A  5), 

As  cos  AD  :  radius  : :  cos  AC  :  cos  DC ; 
hence  (23.5), 

As  cos  AD  :  cos  AB  : :  cos  BC  :  cos  DC. 

Q.  E.  D. 


Art.  71.  The  same  things  being  supposed  as  in  the  last 
article,  the  tangents  of  the  bases  are  to  each  other  as  the 
cosines  of  the  adjacent  angles.     For  (Art.  62), 

As  radius  :  cos  CAB  : :  tan  AC  :  tan  AB ; 


SECTION   III.  r5 

and,  by  inversion, 

As  cos  CAD  :  radius  : :  tan  AD  :  tan  AC ; 

therefore  (22.5), 

As  cos  CAD  :  cos  CAB  :  i  tan  AD  :  tan  AB. 

Q.  E.  D. 


A 


Art.  72.  As  the  sum  of  the  sines  of  any  two  unequal  arcs 
is  to  their  difference,  so  is  the  tangent  of  half  the  sum  of 
those  arcs,  to  the  tangent  of  half  their  difference. 

Let  AB,  AC  be  the  arcs ; 
L  the  centre  of  the  circle ; 
and  AH  the  diameter  passing 
through  A  ;  make  AF=AB; 
join  BF,  and  let  BF  cut  AH 
in  D ;  draw  CE  parallel  to 
BF,andCG  to  AH;  let  CG 
meet  BF  in  I ;  join  GB,  GF 
and  CF.  Then,  since  AF 
=  AB,  FD  =  BD;  and 
HDF  =  HDB ;  hence  BD  is  the  sine  of  AB ;  and  CE,  which 
is  parallel  to  BD,  is  the  sine  of  AC ;  therefore,  FI  is  the  sum, 
and  BI  the  difference,  of  the  sines  of  AB  and  AC. 

Again,  the  arc  CF  is  the  sum,  and  CB  the  difference,  of 
AB  and  AC ;  therefore  the  angle  FGC  is  measured  by  half 
the  sum,  and  BGC  by  half  the  difference,  of  AB  and  AC 
(20.3).  But  the  angle  GIF  =  HDF  (29.1),  and  is  therefore 
a  right  angle ;  consequently,  IF  is  the  tangent  of  CGF,  and 
IB  the  tangent  of  CGB,  to  the  radius  GI ;  therefore,  for  any 
other  radius  (Art.  27,  Cor.), 


c 

\    £ 

I 
7) 

\  A 

^r 

/E   1 

Ug 

As  IF  :  IB  : :  tan  CGF  :  tan  CGB 
i(AB-AC). 


tan  i(AB  +  AC)  :  tan 
Q.  E.  D. 


Art.  73.  The  sum  of  the  cosines  of  two  unequal  arcs  is, 


(£  SPHERICAL  TRIGONOMETRY. 

to  their  difference,  as  the  cotangent  of  half  their  sum  is  to 
the  tangent  of  half  their  difference. 

Retaining  the  construction  of  the  last  article,  it  is  easily 
perceived  that  GI  is  the  sum,  and  IC  the  difference,  of  the 
cosines  of  AB  and  AC ;  also  the  angle  GFI  is  the  complement 
of  CGF;  and  IFC  -  BGC;  hence, 

cos  AB  +  cos  AC  :  cos  AC  —  cos  AB  ::  cotan  ^(AB  +  AC)  : 
tan  J(AB  —  AC).  Q.  E.  D. 

Art.  74.  In  any  oblique  angled  spherical  triangle,  a  per- 
pendicular being  let  fall  from  the  vertex  on  the  base,  i,t  will 
be,  as  the  tangent  of  half  the  base  is  to  the  tangent  of  half 
the  sum  of  the  sides,  so  is  the  tangent  of  half  the  difference 
of  those  sides  to  the  tangent  of  the  distance  between  the 
c  perpendicular  and  mid- 

dle of  the  base. 

Let  ABC  be  the  trian- 
gle, CD  the  perpendicu- 
and  E  the  middle  of 
base.      Then    (Art. 

As  cos  AC  :  cos  BC  : :  cos  AD  :  cos  BD ; 

whence  (E  5), 

As  cos  AC  +  cos  BC  :  cos  BC  —  cos  AC  : :  cos  AD  +  cos  BD  ; 
cos  BD  —  cos  AD ; 

consequently  (Art.  73), 

As  cotan  l(AC  +  BC)  :  tan  J(AC  —  BC)  :  :  cotan  A(AD  + 
DB)  :  tanl(AD  — DB)  ::  cotan  AE  :  tan  ED ; 

and,  alternately, 

As  cotan  i( AC +  BC)  :  cotan  AE  ::  tan  £(AC— CB) :  tan  ED. 
But  (Art.  23.1), 
cotan  i(AC  +  BC)  :  cotan  AE  : :  tan  AE  :  tan  J(AC+BC); 


SECTION  Jill  97 

therefore, 

As  tan  AE  :  tan  £(AC  +  BC)  ::  tan  i(AC  — BC)  :  tan  ED. 

q:  E.  D. 

Art.  75.  In  any  oblique  angled-  spherical  triangle,  a  per- 
pendicular being  let  fall  from  the  vertex  on  the  base,  and  an 
arc  described  bisecting  the  vertical  angle ;  it  will  be,  as  the 
cotangent  of  half  the  sum  of  the  angles  at  the  base  is  to  the 
tangent  of  half  their  difference,  so  is  the  tangent  of  half  the 
vertical  angle  to  the  tangent  of  the  angle  formed  by  the  per- 
pendicular and  the  arc  bisecting  the  vertical  angle. 

Let  ABC  be  the  triangle,  CD  the  perpendicular,  and  CF 
the  arc  bisecting  the  vertical  angle ;  then  (Art.  68), 

As  cos  A  :  cos  B  : :  sin  ACD  :  sin  BCD ; 

hence  (E  5) 

cos  A+cos  B  .  cos  A  —  cos  B  ::  sin  ACD  -f  sin  BCD  :  sin 
ACD  —  sin  BCD ; 

therefore  (Arts.  72,  73) 

As  cotan  J(A  +  B)  :  tan  i(B  —  A)  ::  tan  $(ACD  +  BCD) 
:tan|(ACD  — BCD)  ::  tan  ACF  :  tan  DCF. 

Q.  E.  D. 

Scholium.  From  the  analogies  demonstrated  in  Articles  60, 
62,  63,  65,  66  and  67,  we  may  frequently  determine  the  affec- 
tions of  the  sides  and  angles  of  the  triangles,  by  adverting 
to  the  signs  of  the  terms,  as  explained  in  Art.  24.  Thus,  in 
Art.  60,  the  base  AB  being  always  less  than  a  semicircle,  the 
sin  AB  is  positive;  hence  the  tan  BAC  and  tan  BC  are  both 
positive  or  both  negative;  consequently,  BAC  and  BC  are 
both  less  or  both  more  than  90°.  In  Art.  62,  when  the  angle 
at  the  base  is  acute,  its  cosine  is  positive ;  consequently,  the 
tangents  of  the  hypothenuse  and  base  will  be  both  positive  or 
both  negative ;  therefore  the  arcs  themselves  will  be  both  more 
or  both  less  than  90° ;  that  is,  they  will  be  of  the  same  affec- 
13  i 


08  SPHERICAL   TRIGONOMETRY. 

tion.  But  when  the  angle  at  the  base  is  obtuse,  its  cosine 
will  be  negative ;  and  therefore  the  tangent  of  the  hypothe- 
nuse  and  that  of  the  base  will  be  one  positive,  and  the  other 
negative ;  consequently,  the  arcs  themselves  will  be  of  dif- 
ferent affections.  In  Art.  63,  when  the  vertical  angles  are 
of  the  same  affection,  their  cosines  have  the  same  sign ;  con- 
sequently, the  tangents  of  the  adjacent  sides  will  have  the  same 
sign,  and  will  therefore  be  of  the  same  affection.  The  same 
principles  are  applicable  to  the  other  cases.  The  conclusions 
thus  obtained  are  consonant  to  those  obtained  in  a  different 
manner  in  Arts.  56,  57. 

The  analogies  above  demonstrated  are  sufficient  to  enable 
the  student  to  calculate  the  sides  and  angles  of  spherical  tri- 
angles from  the  usual  data;  yet  there  are  various  useful 
forms,  hereafter  demonstrated,  which  are  applicable  to  par- 
ticular cases.  We  have  also  two  concise  rules,  discovered 
by  Baron  Napier,  the  celebrated  inventor  of  logarithms,  by 
which  the  cases  in  right  angled  spherical  triangles  are  con- 
veniently solved ;  and,  being  easily  remembered,  they  are 
frequently  used  in  practice. 


Art.  76.  In  right  angled  spherical  triangles,  there  are  five 
parts  which  may  have  different  values  assigned  to  them 
without  changing  the  right  angle,  viz. :  the  hypothenuse,  the 
two  sides,  and  the  two  oblique  angles.  Now,  the  sides,  and 
the  complements  of  the  hypothenuse  and  oblique  angles,  are 
called  the  five  circular  parts ;  one  of  which  being  assumed  as 
the  middle  part,  the  two  which  lie  contiguous  to  this  middle 
part  are  called  the  adjacent  extremes ;  and  the  other  two 
are  termed  the  opposite  extremes.     Then  Napier's  rules  are : 

1.  The  rectangle  of  radius  and  the  sine  of  the  middle  part 
is  equal  to  the  rectangle  of  the  tangents  of  the  adjacent 
extremes. 

3.  The  rectangle  of  radius  and  the  sine  of  the  middle 


SECTION   III. 


99 


part  is  equal  to  the  rectangle  of  the  cosines  of  the  opposite 
extremes. 

These  rules  may  be  explained  and  demonstrated  in  the 
following  manner : 

c  Let  ABC  be  the  tri- 

angle, right  angled 
at  B.  Then,  assum- 
ing AB  as  the  mid- 
dle part,  the  side  BC 
and  complement  of 
BAC  are  the  adja- 
cent extremes ;  and 
the  complements  of 
AC  and  ACB  are  the 
opposite  extremes.     Now  (Art.  60), 

As  radius  :  sin  AB  : :  tan  BAC  :  tan  BC ; 
and,  alternately, 

As  radius  :  tan  BAC  : :  sin  AB  :  tan  BC. 
But  (Art.  23), 

As  radius  :  tan  BAC  : :  cotan  BAC  :  radius. 
Hence, 

cotan  BAC  :  radius  : :  sin  AB  :  tan  BC ; 
therefore  (16.6), 

radius  .  sin  AB  =  cotan  BAC  .  tan  BC ; 
which  is  Napier's  first  rule. 

Again  (Art.  58), 

As  radius  :  sin  ACB  : :  sin  AC  :  sin  AB ; 
whence  (16.6), 

radius  .  sin  AB  =  sin  ACB  .  sin  AC ; 
which  is  Napier's  second  rule. 


100  SPHERICAL  TRIGONOMETRY. 

If  we  assume  BC  the  middle  part,  AB  and  the  complement 
of  ACB  become  the  adjacent  extremes;  and  the  complements 
of  BAC  and  AC,  the  opposite  extremes.  Napier's  rules  may- 
then  be  demonstrated  in  that  case  exactly  as  before. 

Assuming  next  the  complement  of  BAC  as  the  middle  part, 
AB  and  the  complement  of  AC  become  adjacent  extremes ; 
and  BC  and  the  complement  of  BCA  opposite  extremes. 
Then  (Art.  62), 

As  radius  :  cos  BAC  : :  tan  AC  :  tan  AB ; 

alternately, 

As  radius  :  tan  AC  : :  cos  BAC  :  tan  AB. 

Hence  (Art.  23), 

As  cotan  AC  :  radius  : :  cos  BAC  :  tan  AB ; 

consequently  (16.6), 

radius  .  cos  BAC  ==  cotan  AC  .  tan  AB ; 

which  is  rule  first. 

Again  (Art.  67), 

As  radius  :  sin  BCA  :  cos  BC  :  cos  BAC ; 
whence, 

radius  .  cos  BAC  =  sin  BCA  .  cos  BC ; 
which  is  rule  second. 

In  the  same  manner,  the  rule  is  demonstrated,  when  the 
complement  of  ACB  is  taken  as  the  middle  part. 

Lastly,  assuming  the  complement  of  AC  as  the  middle  part, 
the  complements  of  BAC  and  BCA  are  the  adjacent  extremes ; 
and  AB,  BC,  the  opposite  extremes.     Then  (Art.  64), 

As  radius  :  cos  AC  : :  tan  ACB  :  cotan  BAC ; 

alternately, 


SECTION    III.  101 

As  radius  :  tan  ACB  : :  cos  AC  :  cotan  BAC ; 
wherefore  (Art.  23), 

As  cotan  ACB  :  radius  : :  cos  AC  :  cotan  BAC ; 
consequently, 

radius  .  cos  AC  =  cotan  ACB  .  cotan  BAC  ; 
which  is  rule  first. 

And  (Art.  65), 

As  radius  :  cos  AB  : :  cos  BC  :  cos  AC ; 
whence, 

radius  .  cos  AC  =  cos  AB  .  cos  BC ; 
which  is  rule  second. 


The  following  table  exhibits  the  different  cases,  and  the 
equations  arising  from  Napier's  rules : 


o 

S 
■ 

1 

Middle 
part. 

Adjacent  Ex- 
tremes. 

Opposite  Ex- 
tremes. 

Equations. 

AB 

Comp  BAC 
BC 

Comp  ACB 
Comp  AC 

a    ■    kt>      CcotBAC.tan  BC 
rad.sinAB=£8inACBt8inAC 

•2 

BC 

CompBCA 
AB 

CompBAC 
Comp  AC 

j    •  on      Ccot  BCA.tan  AB 

rad.sinBC= -?    .     t>  a  r<  „:     a  n 

£  sin  fcsAC.sin  AC 

3 

Comp  BAC  iComPAC 

Comp ACB 
BC 

j        -d  a  n      C  cot  AC. tan  AB 
rad.cosBAC=^8inACBcoBBC 

4 

CompBCA;001^^ 

Comp  BAC 
AB 

j       t)/-.  a      Ccot  AC. tan  BC 
rad.cosBCA=^sinBACcosAB 

5 

n          a /-i   ICompBAC 
Comp  AC   !Com£BCA 

AB 
BC 

.        ,n      Ccot  BAC.cotBCA 
rad.cosAC=-?          AT}  m  m  ^^ 
£  cos  AB.cos  BC 

When  any  two  of  these  circular  parts  are  given  to  find  a 
third,  we  must  assume  such  one  to  be  the  middle  part  as  will 
make  the  other  two  either  both  adjacent  or  both  opposite 
extremes. 

The  following  practical  examples  will  serve  to  exercise  the 
preceding  theory : 

14 


102 


SPHERICAL  TRIGONOMETRY. 


Example  1.  In  the 
spherical  triangle 
ABC,  right  angled 
at  B,  given  the  side 
AC  52°  15',  and  the 
angle  A  23°  28',  to 
find  the  other  sides 
and  the  remaining 
angle. 

The  perpendicular 
BC  may  be  found  by  Art.  58 ;  the  side  AB,  by  Art.  62 ;  and 
the  angle  C,  by  Art.  64.  Or,  using  Napier's  circular  parts, 
we  find  BC  by  the  second  equation,  case  2,  in  the  foregoing 
table ;  AB  by  the  first  equation,  case  3 ;  and  the  angle  C  by 
the  first  equation,  case  5.  The  results  are,  BC  18°  21'  9" ; 
AB  49°  49'  57";  C  75°  6'  58". 


Ex.  2.  Given,  the  base  AB  61°  25',  and  the  adjacent  angle 
A  32°  45',  to  determine  the  rest. 

The  hypothenuse  AC  may  be  found  by  Art.  62 ;  the  per- 
pendicular BC  by  Art.  60 ;  and  the  angle  C  by  Art.  67 ;  or 
by  cases  3, 1,  4  of  the  circular  parts. 

The  results  are,  AC  65°  22'  52" ;  BC  29°  27'  32" ;  and  C 

75°. 

Ex.  3.  Given,  the  base  AB  75°  28',  and  the  perpendicular 
BC  41°  15',  to  find  the  rest. 

The  results  are,  AC  79°  7'  30";  A  42°  10'  32";  C  80° 
18'  1". 


Ex.  4.  Given,  the  angle  A  23°  28'  30",  and  angle  C  75°  22', 
to  find  the  rest. 

The  results  are,  *$  53°  2'  36";  AB  50°  38'  22";  BC  18° 
33'  40". 


SECTION    III.  103 

Ex.  5.  In  the  oblique 
angled  triangle  ABC, 
given  the  sides  AB  70°, 
AC  58°,  and  the  angle 
CAB  52°  30',  to  find  the 
rest. 

Suppose  the  arc  CD  of 
a  great  circle  at  right  angles  to  AB  to  pass  through  C ;  then 
the  given  triangle  will  be  divided  into  two  right  angled  ones, 
ADC  and  BDC. 

In  the  triangle  ADC,  the  side  AD  and  angle  ACD  may  be 
computed  by  Arts.  62  and  64.  Hence  BD  is  known.  Then 
the  angle  B,  the  side  BC,  and  the  angle  BCD,  may  be  found 
by  Arts.  61,  66  and  69. 

Results :  B  64°  28' ;  BC  48°  12'  46" ;  ACB  91°  0'  21". 

Ex.  6.  In  the  spherical  triangle  ABC,  given  the  angle  BAC 
50°  15',  ACB  92°,  and  side  AC  57°  30',  to  find  the  rest. 

The  arc  CD  being  made  perpendicular  to  AB,  the  side  AD 
and  angle  ACD  may  be  found  as  in  the  last  example ;  whence 
BD,  BC,  and  the  angle  at  B,  may  be  computed  by  Arts.  69, 
63  and  68. 

Results:  BA  69°  25*  2";  BC  46°  4'  16";  ABC  64°  12'  16". 

Ex.  7.  In  the  triangle  ABC,  given  AB  71°  30' ;  AC  59° 
20';  BC  50°  10' ;  to  find  the  angles. 

Drawing  CD  at  right  angles  to  AB,  and  taking  AE  =  ■JAB, 
the  arc  ED  is  found  by  Art.  74 ;  from  which  AD  and  BD 
become  known ;  and  thence  the  angles  may  be  found  by  Arts. 
62  and  59. 

Results:  BAC  54°  3'  51";  ABC  65°  5' 4";  ACB  90°  48' 
47". 

Ex.  8.  In  the  triangle  ABC,  given  the  angle  BAC  51° ; 
ABC  58°;  and  ACB  110°;  to  find  the  sides. 


104 


SPHERICAL   TRIGONOMETRY. 


Using  the  construction  of  the  last  example,  and  describing 
CF  so  as  to  divide  the  vertical  angle  into  two  equal  angles, 
the  angle  FCD  may  be  found  by  Art.  75 ;  whence  the  angles 
ACD  and  BCD  become  known ;  and  thence  the  sides  AC,  BC 
may  be  determined  by  Art.  64 ;  and  AD,  BD,  by  Art.  67. 

Results  :  AC  64°  28'  31" ;  BC  55°  47'  13"  ;  AB  90°  44'  26". 


Art.  77.  It  has  been  already  mentioned  that  there  are 
various  useful  forms  which  are  applicable  to  particular  cases. 
By  means  of  these,  the  necessity  of  dividing  an  oblique 
angled  triangle  into  two  right  angled  ones,  is  always 
obviated.  Of  these  forms,  the  following  are  the  most  im- 
portant. They  are  investigated  most  conveniently  by 
algebra. 

c  Let  ABC  be  a  spherical 

triangle ;  CD  a  perpendi- 
cular upon  AB;  and,  to 
accommodate  the  expres- 
t  sions  to  the  language  of 
algebra,  let  the  capital 
letters  A,  B,  C  denote  the 
angles,  and  the  small  letters  a,  b,  c,  the  opposite  sides ;  the 
segments  AD,  BD,  being  represented  by  d,  e,  and  the  oppo- 
site angles  ACD  and  BCD  by  D  and  E  respectively.  In 
these  investigations,  the  radius  is  taken  =  1. 


Now  (Art.  62), 

1  :  cos  A 


tan  b  :  tan  d. 


But  (Art.  23), 


sin                      sin  d  sin  b 

■=  tan  .-.  r  =  cos  A. 


cos  d 


cos  b 


cosin 
Again  (Art.  66). 

cos  b  :  cos  a  : :  cos  d  :  cos  (c  —  d) ; 
wherefore, 


SECTION    III.  105 

cos  a       cos  (c — d)      ,-     n„  _         ~,cos  c.cos  a7 + sine. sin d 

r  = --rJ-=  (Art.  37,  Form.  2) j 

cos  o  cos  a         x  '  cos  a 

sin  d                             .    .        sin  b 
—  cos  c  +  sine.-- — ;  =  cos  c  +  cos  A.sinc. r;  by  putting 

cos  a  cos  o     J  l  ° 

.  sin  b  .  sin  J 

cos  A. 1  instead  ot .. 

cos  b  cos  a 

Clearing  this  equation  of  fractions, 

cos  a  =  cos  c.cos  b  +  cos  A.sin  c.sin  b,  (1) 

By  Art.  64, 

_  _        ;  •'■-  "  •■'  i^  sin  B      cos  E 

As  1  :  cos  a  : :  tan  B  :  cot  E  : :  (Art.  23) ~   :  ~ — =cr  •"• 

v  '  cos  B      sin  E 

cos  E        .         sin  B 


COSG. 


sin  E  'cos  B' 

By  Art.  68, 

As  cos  B  :  cos  A  : :  sin  E  :  sin  (C  —  E)  .-. 

cos  A      sin  (C — E)     ,k       _,_  _,  fc%  sin  C.cos  E — cos  C.sin  E 

^  = r^— =(Art.  37,  F.  1) .— -^ 

cos  B  sin  E  v  '  sin  E 

.       cos  E  sin  B 

=  sin  C- — ts  —  cos  C.  —  cos  a.sin  C. ^  —  cos  G ;  substi- 

sin  E  cos  B 

cos  E 
tuting  for- — ^.     Then,  clearing  of  fractions, 
6      sin  E  6  ■ 

cos  A  =  cos  a. sin  C.sin  B  —  cos  C.cos  B.  (2) 


By  Art.  62, 

As  1  :  cos  B  : :  tan  a  :  tan  e  : :  (Art.  23, 4)  cotan  e  :  cotan  a 

_  _  cos  e 

.*.  cotan  a  =  cos  B.cotan  e  =>  cos  B.— ; 

sin  e 

.       .  cotan  a       cos  e 

wherefore,  =r-  =  — « • 

cos  B         sin  e 

14 


100  SPHERICAL  TRIGONOMETRY. 

But  (Art.  61), 

As  tan  A  :  tan  B  : :  sin  e  :  sin  (c  —  e) ; 

therefore, 

tan  B      sin  (c  —  e)        /k       -V     _   ,x   sin  c.cos  e. — cosc.sine 

; T  = ^ =  (Art.  37,  F.  1) : 

tan  A  sin  e  v  '  sin  e 

cos  e  .       cotan  a 

=  sin  c- cos  c  =  sin  c. ^ cos  Ct 

sin  e  cos  J3 

Consequently,  by  clearing  of  fractions, 

cos  B.tan  B  =  cotan  a.sin  c.tan  A  —  cos  c.cos  B.tan  A. 

Now, 

cos  B.tan  B=sin  B  (Art.  23)  and  r  =  cotan  A. 

tan  A. 

sin  B.cotan  A+cos  c.cos  B  ._. 

Hence  cotan  a  = : .  (o) 

sine       . 

By  Art.  64, 

As  1  :  cos  b  : :  tan  D  :  cotan  A=cos  fr.tan  D=cos  b. ^ 

cos  D 

By  Art.  63, 

As  tan  a  :  tan  b  : :  cos  D  :  cos  (C  —  D) ; 

tan  b       cos  (C  —  D)  ow  _x 

.-. ; =  * — ^r--=  (Art.  37,  Form.  2)  cos  C 

tan  a  cos  D  v  [ 

.        sin  D      sin  D      tan  ft.cotan  a  —  cos  C 

+  sin  C. ^  .-. =p:  = ; — ~ ' 

cos  D      cos  D  sm  U 

Consequently, 

sin  fr.cotan  a —  cos  C.cos  & 

cotan  A  == : — ~ .  (4) 

sm  C  v  ' 

These  forms  are  not  suited  to  logarithmic  computations ; 
but  they  are  useful  in  the  investigation  of  othor  equations,  to 
which  logarithms  are  conveniently  applied. 


SECTION   III.  107 

From  Form.  1,  above  given,  we  find,  by  transposition  and 
division, 

.       cos  a  —  cos  c.cos  b 

cos  A  = - 1 — ; 

sin  c.sin  b 

hence, 

cos  a  —  cos  c.cos  b 


1  —  cos  A  =  1 


sin  c.sin  b 


cos  c.cos  b  +  sin  c.sin  b  —  cos  a       ,:,       „     „ 

1 r-t =  (Art.  37,  Form.  2) 

sin  c.sin  b  v  ' 

cos  (c  —  b)  —  cos  a 
sin  c.sin  b 

But  (Art.  36,  Form.  6), 

1  —  cos  A  =  2  sin9  -JA ; 
and  (Art.  37,  Form.  13), 

cos  (c  —  b)  —  cos  a  =  2  sin  %(a  +  c  —  b).sm  ±(a  +  b  —  c) 


.-.  sin5 


u_sin  i(a  +  c — 6) .sin  £(a  +  b — c) 

■syA  — r ; r^ 


sin  c.sin  b 


(5) 


.      .     c        ,,  ..  .       cos  a  —  cos  c.cos  b 

Again,  trom  the  equation  cos  A  = .   we 

sm  c.sin  b 

have 

cos  a  —  cos  c.cos  b 


1  +  cos  A  =  1  +■ 


sin  c.sin  b 


cos  a  —  cos  c.cos  b  +  sin  c.sin  b 

"Iran"*-  "=  (Art.  37,  Form.  2) 

cos  a  —  cos  (c  +  b) 
sin  c.sin  b 

But  (Art.  36,  Form.  8) 

1  +  cos  A  =  2  cos8  £A ; 


108  SPHERICAL   TRIGONOMETRY. 

*  sin  c.sin  b  v  ' 

Again : 

tan*  i  A  -  sin3  ^A  -  si"  l(a+c—b)  sin  j(a +b— c) 

*         coss|A      sin  i(c  +  b  +  a)  sin  i(c  +  b— a)'       ™ 

Equations  5,  6  and  7  furnish  convenient  expressions  for 
finding  an  angle,  when  the  three  sides  are  given. 
From  Form.  2, 

cos  A  +  cos  C.cos  B 

cos  a  = . — ~r--. — s ; 

sin  C.sin  B 

wherefore, 

_      cos  A+cos  C.cos  B 

1 COS  G=  1 . — 77-: — =i = 

sin  C.sin  B 

sin  C.sin  B — cos  C.cos  B — cos  A 
sin  C.sin  B 

— (cos  C.cos  B — sin  C.sin  B) — cos  A  _  — cos  (B  +  C) — cos  A 
sin  C.sin  B  sin  C.sin  B 

cos  (B  +  C)  +  cos  A 

= •    n   •    p =  (Art-  37,  Form.  12) 

sin  C.sin  B  v  ' 

—  2  cos  |(B  +  C  +  A).cos  fr(B  +  C  — A) 
sin  B.sin  C 

But  (Art.  36,  Form.  6), 

1  —  cos  a  =  2  sin3  \a  ; 

Wherefore, 

«*£■  -cosKB-fC+Aj^KB+C-A), 

J  sin  B.sin  C  v  ' 


*  Since  the  three  angles  of  a  spherical  triangle  are  together  greater  than 
180°,  but  less  than  540°  (Art.  54,  Cor.),  cos  i(B+C  +  A)  will  always  be 
a  negative  quantity,  and  consequently  — cosine  a  positive  quantity. 


SECTION   III.        •  109 

Again : 

_  _      cos  A+cos  C.cosB 

1  -f  cos  a=  H -. — 7^—: — ts = 

sin  C.sin  B 

cos  C.cos  B  -f  sin  C.sin  B  +  cos  A  _  cos  (C  —  B)+cos  A 
sin  C.sin  B  sin  C.sin  B 

(Art.  37,  Form.  2). 

But  (Art.  36,  Form.  8), 

1  +  cos  a  =  2  cos3  \a ; 

and  (Art.  37,  Form.  12) 

cos  (C  — B)+cos  A=2  cos£(A  +  C  — B).cos  J(A+B  — C); 

whence, 

cos*  ifl      cos|(A+C-B).cosKA+B~C) 

cos  ?a-  sin  C.sin  B  *  W 


Further : 
♦    ,i       sin2  \a     -cosKB+C+A).cosj(B+C--A) 
tan  ^a  ~~coss  \a  ~  cos  i(A  +  C  — B).cos  |(A+B— C)'       ^1U> 

Equations  8,  9  and  10  may  be  conveniently  used  for  de- 
termining the  sides,  when  all  the  angles  are  given. 

By  Art.  59, 

As  sin  b  :  sin  a  : :  sin  B  :  sin  A; 
wherefore  (E.  5), 

sin  b  +  sin  a  :  sin  b  —  sin  a  ::  sin  B  + sin  A  :  sin  B  —  sin  A; 
consequently,  by  Art.  72, 

As  tan  J(6+a) :  tan  l(b— a) : :  tan  £(B  +  A)  :  tan  £(B— A).  (N) 
wherefore, 

ta„KB-A)=ta„KB+A).^||=§  (P) 

15 


110  SPHERICAL   TRIGONOMETRY. 

Again  (Art.  68), 

As  cos  B  :  cos  A  : :  sin  E  :  sin  D ; 
hence  (E.  5), 

As  cos  B  +  cos  A :  cos  A — cos  B  : :  sin  E  +  sin  D :  sin  D — sin  E ; 
consequently  (Arts.  73.72), 

As  cotan  J(B+ A)  :  tan  J(B— A)  : :  tan  JC  :  tan  i(D— E) ; 
whence 

tan  .(D-E)=ta„  .C.-^i|^=  (Art. 23.4) 

tan  JC.tan  * (B  —  A).tan  \ (B  +  A). 

By  Art.  63, 

As  tan  b  :  tan  a  : :  cos  E  :  cos  D ; 

wherefore  (E.  5), 

As  tan  J  +  tan  a  :  tan  b — tan  a  : :  cos  E  +  cos  D  :  cos  E — cos  D. 

But  (Art.  37,  equation  8), 

,    ,  sin  (b  ±  a) 

tan  b  db  tan  a  =  2 r  ; 

cos  a.cos  o 

consequently, 

sin  (a  +  b)  :  sin  (b  —  a)  : :  cotan  JC  :  tan  J(D  —  E) ; 

wherefore, 

tan  i(D  — E)=cotan^C.S?n  ,(f  "7^=  (Art-  36>  eq-  3) 

JX  '  sin  (b  +  a)      v  '     ' 

2  cos  j(b  —  a). sin  ±(b —  a) 
cotan  nC-2-—-T(a  +  b).sm  ±(a  +  b)' 

Equating  these  values  of  tan  £(D —  E), 

tan  JC.tan  J(B  —  A).tan  i(B  +  A)  = 

cos  h (b  —  q).sin  |  (b  —  a) 
cotan  u^'cos  i (a  +  ^.sjn  J(a  +  &)" 


SECTION   III.  Ill 

In  this  equation,  substitute  for  tan  J(B  —  A),  its  value  given 
in  equation  P ;  then, 

tan  JCton'  >(B  +  A)'a"^~a|= 
2V  '.  tan  l(b  +  a) 

cos  l(b  —  a). sin  Mb  —  a) 
C0tan  iUcos  i(a  +  b).s'm  J(a  +  b)' 

,  i/p  ,.  A\  _  cotan  2C»cos  *(&— oQ.sin  j(&— a).tan  j(&+a) 
tan  3^b+a;-   tan  iC.COs  |(«+^).sin  |(a+6).tan  |(^— a)  * 

Now  f  Art.  22.1V 


sin        1    _  tan 
tan  '  *  cos      sin 


Hence, 
tn«i(J 

Now  (Art.  23.1), 

and  (Art.  23.4), 

1        cotan 

cotan  =  - —  .*.  — =  cotan3. 

tan        tan 

Our  equation  is  therefore  reducible  to  this : 

tan'  i(B+A)=cotan'  iC.cos3  ^~  «>  ■ 
^X  '  z      cos-8  l(a+b) 

consequently, 

tan*(B+A)=cotan^g=|.  (11) 

From  equation  P, 

i/t>       a\  incosK^  —  a). ten  Mb —  a) 

tan  |(B  — A)=cotan-lC. 177— ; — (-7 — -tj—, — C  = 

2V  y  2    cos  l(b  +  a). tan  i(o  -f  a) 

,>_  sin  l(&  —  a) 
cotan  ifi,    A  ^   v  (12) 

sin  i(b  +  a)  K     ' 

From  proportion  N, 
As  tan  h(B  +  A)  :  tan  i(B— A)  : :  tan  i(b  +  a)  :  tan  £(&— a) 


112  SPHERICAL   TRIGONOMETRY. 

x  tani(B  —  A)  ^ 

...  tan  £(&-a)=tan  W  +  fl)-^  h\B  +  Aj-  (0) 

By  Art.  66, 

As  cos  a  :  cos  5  : :  cos  e  :  cos  d ; 
therefore  (E.  5  and  Art.  73), 

As  cot  z(b+a)  :  tan  i(b  —  a)  : :  cot  £c  :  tan  i(d —  e)  ; 
whence, 

tan  Ud  —  e)=cot  ic.-^^r-^T  =  (Art-  23,  ecl-  4) 
v  '  cotan  i(6  +  a)       v 

cot  ic.tan  1(6  -f  a). tan  £(&  — a). 

Again  (Art.  61), 

As  tan  B  :  tan  A  : :  sin  d  :  sin  e ; 

therefore  (E.  5  and  Art.  72), 

As  tan  B  +  tan  A  :  tan  B — tan  A  : :  tan  \c  :  tan  h(d — e) ; 

.    tan  B  —  tan  A     ,  .    .■  ow  _v 

.-.  tan  Ud  —  e)=tan  £c. s— - r=(Art.  37,  eq.  8) 

v  '  tan  B  +  tan  A     v 

,    sin  (B  —  A)       ,  .       nrt         _. 
tan^-sinlB— Al  =  (Art-36^e^3) 

1    sini(B  —  A).cosi(B  — A) 
tan  3jc.gin  ^R  +  A^cog  i(B  +  Ay 

Equating  these  values  of  tan  h(d  —  e),  and  substituting 
for  tan  h(b  —  a)  its  value  in  equation  Q; 

,  ,        „  tan  l(B  — A) 
cot  icm*  ,(b  +  a)  -J^g-j^  = 

sini(B  — A)    cos  KB  — A)  , /7       r 

tan  he  - — .^ '  Vn  •  — nfopr  ,    Ai  •*•  tan*  5(6  + «)= 
sin  i(B  -f  A)   cos  A(B  -f  A)  v        ' 


SECTION   III.  113 

tan  Ic.sin  |(B  —  A).cos  i(B  —  A).tan  j(B+A) 
cotk.sin  i(B  +  A).cos  £(B  +  A).tan  i(B  —  A)    ~ 

cos»  KB  — A) 
tan9  \c- 


cos3£(B  +  A)  * 
consequently, 


„       i  i    cosi(B  — A) 

tanK^  +  «)=tanic.^gT-A-J.  (13) 

From  equation  Q, 

ta„i(J-a)=ta„ic.S4^|^.  (14) 

Equations  11  and  12  may  be  used  when  two  sides  and  the 
included  angle  are  given  to  find  the  other  angles ;  and  equa- 
tions 13  and  14  when  two  angles  and  the  side  between  them 
are  given  to  find  the  other  sides.* 

From  these  four  last  equations,  the  following  are  derived 
by  a  very  simple  process : 

cota„|C=tani(B+A).£|^.  (15) 

cotaniC=tan|(B-A).^||^.  (16) 

tanic=tan^+a).^|||^^.  (17) 

tan  ic=tan  J(A  -  a).^  l(B_S^y  (18) 

A  few  examples  are  given  to  exercise  these  equations. 


*  The  discovery  of  these  four  equations  is  attributed  to  Baron  Napier. 
15  K  * 


114 


SPHERICAL  TRIGONOMETRY. 
Z 


Ex.  1.  In  the  spheri- 
cal triangle  ABZ,  given 
AZ=54°  10',  BZ  m  39° 
25',  AB  =  72°  36',  in 
ZA  produced  Aa  =  2', 
and  in  ZB,  B6  =  35',  to 
find  ab. 


First,  with  the  three   sides,  find   the   angle  Z,  by  equa- 
tion 5. 


ZA    54°  10' 

cosec     .0911273 

ZB    39°  25' 

cosec    .1972569 

AB    72°  36' 

2)166°  11' 

S    83°    5'  30" 

S— ZA    28°  55'  30" 

sine  9.6845440 

S— ZB    43°  40'  30" 

sine  9.8392057 

2)19.8121339 

\Z    53°  39'  32" 

sine  9.9060669 

BZA  107°  19'    4" 

Then,  by  equations  11  and  12, 
l(Za  +  Zb)    46°  31'  sec    .1623210    cosec    .1393179 


|(Za— Zb) 

7°  41' 

cos  9.9960834    sin 

9.1261246 

PZA 

53°  39'  32" 
46°  39'  15" 

cot  9.8666880    cot 

9.8666880 

h(b+a) 

tan  10.0250924 

i(b~a) 

7°  43'  12" 

9.1321305 

Zba 

54°  22'  27" 

SECTION 

III. 

By  Art.  58, 

Zba 

54°  22'  27" 

cosec    .0899959 

bZa 

107°  19'    4" 

sin      9.9798526 

Za 

54°  12' 

72°  17'    8" 

sin      9.9090550 

ab 

sin      9.9789035 

Or  the  side  ab 

may  be  found  by  equation  18. 

i(b-a) 

7°  43'  12" 

cosec     .8718203 

i(b  +  a) 

46°  39'  15" 

sine    9.8616681 

l(Za—Zb) 

7°  41' 
36°    8'  33" 

tan    9.1300413 

\ab 

tan    9.8635297 

ab 

72°  17'    6" 

115 


This  is  the  direct  solution  of  the  celebrated  problem  of 
clearing  the  observed  distance  between  the  moon  and  the 
sun,  or  a  star,  from  the  effect  of  parallax  and  refraction. 

c  Ex.  2.  In  the  spherical  tri- 

angle ABC,  given  AC  46° 
18',  AB  100°  26',  and  the 
angle  A  39°  50',  to  find  the 
rest.  Result:  ACB  136° 0' 
54";  ABC  30°  41'  54"- 
BC  65°  6'  34". 

Ex.  3.  In  the  triangle  ABC,  given  AB  112°  56',  BAC  40° 
16',  ABC  54°  20',  to  find  the  rest. 

Result:  AC  79°  44'  58";  BC  51°  31'  30";  ACB  130°  30' 
20". 

Ex.  4.  Given,  the  side  AB  96°  12',  AC  57°  16',  BC  49°  8', 
to  find  the  angles. 


116  SPHERICAL  TRIGONOMETRY. 

Result :  BAC  31°  32'  42" ;  ABC  35°  35'  15" ;  ACB  136° 

32'  48". 

Ex.  5.  Given,  the  angle  BAC  50°,  ABC  60°,  ACB  85°,  to 
find  the  sides. 

Result:  AB  51°  59'  16";  AC  43°  13'  48";  BC  37°  17 
26". 


SECTION  IV. 


CONIC  SECTIONS. 

Article  78.  Definition  1.  If,  from  a  point  in  the  circum- 
ference of  a  circle,  a  right  line  be  drawn  to  pass  through  a 
fixed  point  which  is  not  in  the  plane  of  that  circle,  and  then 
caused  to  revolve  round  that  fixed  point  so  as  to  describe  the 
whole  circumference  of  the  circle;  the  curve  surface,  de- 
scribed by  this  revolving  line,  is  called  a  conical  surface; 
and  the  solid  included  between  this  curve  surface  and  the 
generating  circle,  is  called  a  cone. 

Def.  2.  The  circle  described  by  the  revolving  line  is  called 
the  base,  and  the  fixed  point  the  vertex,  of  the  cone. 

Def.  3.  The  straight  line  drawn  from  the  vertex  to  the 
centre  of  the  base,  is  called  the  axis  of  the  cone. 

Def  4.  When  the  axis  is  at  right  angles  to  the  plane  of  the 
base,  the  cone  is  called  a  right  cone ;  but  when  the  axis  i3 
oblique  to  that  plane,  the  solid  is  termed  a  scalene  cone. 

As  the  line  which,  by  its  revolution,  describes  the  conical 
surface,  maybe  indefinitely  extended,  two  cones  having  a 
common  vertex,  and  equal  solid  angles  at  the  vertex,  may  be 
generated  by  the  same  revolution. 

Art.  79.  Let  the  cone  ABCD  be  cut  by  a  plane  which 

passes  through  its  vertex  A,  and  cuts  the  base  in  the  right 

line  BC ;  the  common  section  of  this  plane  with  the  surface 

(117; 
16 


118 


SECTION  IV. 


of  the  cone,  will  be  a  triangle.  The 
common  section  of  the  base  and  cutting 
plane  is  a  right  line  (3.2  sup.) ;  and  the 
right  lines  drawn  from  B  and  C  to  the 
vertex,  are  in  the  cutting  plane  (2.2 
sup.) ;  and  those  lines  correspond  to  the 
position  of  the  revolving  line  when  it 
passes  through  B  and  C ;  they  are  there- 
fore in  the  conical  surface. 

Q.  E.  D. 


Art.  80.  Let  the  cone  ABC  be  cut  by  a  plane  which  is 
parallel  to  the  plane  of  the  base ;  then  the  section  of  this 
cutting  plane  with  the  conical  surface,  is  a  circle  whose 
centre  is  in  the  axis  of  the  cone. 

Let  AF  be  the  axis  of  the  cone ; 
DLE  the  cutting  plane.  In  the  cir- 
cumference of  the  base  take  any  point 
K ;  join  FK ;  and  through  AF,  FK, 
suppose  a  plane  to  pass,  cutting  the 
conical  surface  in  AK,  and  the  cut- 
ting plane  in  GH ;  then  (Art.  79,  and 
3.2  sup.)  AK  and  GH  are  right  lines. 
Let  also  another  plane  ABC  pass 
through  the  axis ;  its  section  with  the 
base  will  be  a  diameter,.because  F  is  the  centre  of  the  circle ; 
and  the  section  of  this  plane  with  the  conical  surface  is  a  tri- 
angle (Art.  79).  Take  BC  and  DE,  the  sections  of  this  plane 
with  the  parallel  planes  BCK  and  DLE ;  then  (14.2  sup.)  DE 
and  GH  are  respectively  parallel  to  BC,  FK ;  consequently, 

As  AF  :  AG  : :  BF  :  DG  : :  FC  :  GE  : :  FK  :  GH. 

But  BF,  FC  and  FK  are  all  equal;  therefore,  DG,  GE  and  GH 
are  also  equal ;  consequently  (9.3),  DLEH  is  a  circle  whose 
centre  is  G.  Q.  E.  D. 


CONIC    SECTIONS. 


11<> 


Art.  81.  Let  AB,  DE,  two  lines  at  right  angles  to  each 

other,  such  that  AD.DB  =±  DE2 ; 
then  a  semicircle  described  on 
the  diameter  AB  will  pass 
through  the  point  E. 

Bisect  AB  in  C ;  then,  since 
AD.DB  =  DE3, 

AD.DB  +  CD3  =  DE3  +  CD2; 

that  is  (5.2  and  47.1),  CB3  =  CEa  .-.  CB  £  CE.  Conse- 
quently, a  circle  described  from  the  centre  C,  with  the  radius 
CB,  will  pass  through  E.  Q.  E.  D. 


M   ,~+ 


Art.  82.  Let  ABLC  be 
a  scalene  cone;  ABC  the 
triangle  formed  by  the  sec- 
tion of  the  conical  surface 
with  a  plane  which  passes 
through  the  axis,  and  stands 
at  right  angles  to  the  plane 
of  the  base ;  and  let  another 
plane  GHK,  at  right  angles 
to  the  plane  ABC,  cut  that 
plane  in  GK,  making  the 
angle  AGK  =  ACB,  and 
AKG  =  ABC;*  then  the 
plane  GHK  cuts  the  coni- 
cal surface  in  a  circle. 

In  the  section  of  the  cut- 
ting plane  and  conical  sur- 
face, take  any  point  H; 
through  H  let  a  plane 
DHE      pass,     parallel     to 


the  base  of  the  cone,  cutting  the  planes  GHK  and  ABC  in 
the  lines  HF  and  DE  respectively.     Then,  since  the  plane 


*  This  is  called  a  sub-contrary  section. 


120 


SECTION   IV. 


DHE  is  parallel  to  the  plane  of  the  base,  it  is  at  right  angles 
to  the  plane  ABC  (15.2  sup.).  But  the  plane  GHK  is  at 
right  angles  to  the  same  plane ;  therefore  the  common  section 
HF  is  at  right  angles  to  the  plane  ABC  (18.2  sup.),  and  con- 
sequently to  the  lines  DE  and  GK  in  that  plane  (Def.  1,  2 
sup.) 

Now,  the  angle  GDP  being  =  FKE ;  and  DFG  =  KFE 
(15.1) ;  the  triangles  GDF,  EKF,  are  similar  ;  therefore, 

As  DF  :  FG  : :  FK  :  FE  (4.6) 

consequently,  DF.FE=GF.FK  (16.6).  But  DHE  is  a  circle 
(Art.  80) ;  therefore,  DF.FE  =  HF3  (35.3).  Hence,  GF.FK 
==  HF2 ;  and  consequently  GHK  is  a  circle  (Art.  81),  whose 
diameter  is  GH.  Q.  E.  D. 


Art.  83.  Let  ABC  be 
a  triangle  formed  by  the 
section  of  a  cone  with 
a  plane  passing  through 
its  axis  at  right  angles 
to  the  plane  of  its  base ; 
and  let  another  plane 
DFE,  cutting  the  cone, 
be  at  right  angles  to  the 
plane  of  the  triangle, 
and  so  situated  that 
FG,  the  common  section 
of  these  planes,  shall  be 
parallel  to  AC,  the  op- 
posite side  of  the  trian- 
gle ;  then  the  common  section  of  the  plane  DFE  with  the 
conical  surface,  is  a  curve  called  a  •q^gboly, ;  the  general 
property  of  which  this  article  is  intended  to  exhibit. 

In  this  curve  take  any  point  H,  and  through  II  let  a  plane, 
parallel  to  the  base  of  the  cone,  be  passed ;  and  let  this  plane 
cut  the  plane  of  ABC  in  the  line  LM,  and  the  plane  DFE  in 


CONIC    SECTIONS. 


121 


HK.  Then,  because  the  planes  LHM  and  DFE  are  at  right 
angles  to  ABC,  their  common  section  HK  is  at  right  angles 
to  the  same  plane  (18.2  sup.) ;  it  is  therefore  at  right  angles 
to  FG  and  LM.  Now,  the  section  of  the  plane  LHM  with 
the  conical  surface,  is  a  circle  (Art.  80) ;  wherefore,  LK.KM 
=  HK3  (35.3).  In  like  manner,  BG.GC  ==  DG3.  The  plane 
LHM  being  parallel  to  the  base,  LK  is  parallel  to  BG  (14.2 
sup.) ;  therefore, 
As  FG  :  FK  : :  BG  :  LK  ::  (1.6)  BG.GC  :  LK.KM  : : 

DG3  :  HK3. 
Def.  5.  The  line  FG  is  called  the  axis,  and  F  the  vertex, 
of  the  parabola ;  any  segment  of  the  axis  FK,  reckoned  from 
the  vertex,  is  called  an  abscissa  ;  and  a  perpendicular  KH,  on 
the  axis,  is  called  an  ordinate. 

The  demonstration  in  this  article,  therefore,  shows  that 
any  two  abscissas  are  to  each  other  as  the  squares  of  the 
corresponding  ordinates. 

A  Art.  84.  Let  ABC  be  a  trian- 

gle, formed  by  the  common  sec- 
tion of  a  cone  and  a  plane  through 
its  axis,  at  right  angles  to  the 
plane  of  the  ^ase ;  DIEF  the  com- 
mon section  of  the  conical  sur- 
face, and  a  plane  which  is  at 
right  angles  to  the  plane  of  the 
triangle  ABC,  passing  through 
its  opposite  sides,  but  neither  pa- 
rallel to  the  base,  nor  sub-con- 
trarily  situated ;  the  curve  DIEF 
is  called  an  ellipsp :  the  general 
property  of  which  this  article  is 
designed  to  explain. 

Bisect  DF,  the   common  sec- 
tion of  DIEF  and   ABC,  in  L ; 
take  any  other  point  K  in  DF ; 
and    through  L  and  K  let  the 
16  l 


D 


kO 


M  ( 

~~\iA 

\v\N 

iXLk^ 

I\ 

P  I 

"^7^° 

VT 

Kvy 

122  SECTION    IV. 

planes  MIN  and  PEO  pass  parallel  to  tho  base  of  the  cone 
cutting  ABC  in  MN,  and  PO,  and  DIEF  in  LI  and  KE. 
Through  D  and  F  draw  the  lines  DG  and  HF  parallel  to  BC. 
Now,  since  (15.2  sup.)  the  planes  MIN  and  PEO  are  at 
right  angles  to  the  plane  ABC;  LI  and  KE,  the  common 
intersections  of  these  planes  and  the  plane  DIEF,  are  at  right 
angles  to  the  plane  ABC  (18.2  sup.);  and  consequently  to  the 
lines  MN,PO  and  DF  in  that  plane  (Def.  1.2  sup).  Also  the 
common  sections  of  the  planes  MIN  and  PEO,  and  the  conical 
surface,  are  circles,  of  which  MN  and  PO  are  diameters  (Art. 
80).  Therefore,  ML.LN  =  LP,  and  PK.KO  =  KE2.  Now, 
the  lines  MN  and  PO  are  parallel  to  BC  (14.2  sup).  Hence, 
by  similar  triangles, 


AsDL 

:  DK  :: 

ML 

:PK; 

and 

AsLF 

:  KF  :: 

:  LN 

:  KO; 

therefore 

(23.6), 

As  DL.LF 

:  DK.KF 

: :  ML.LN  :  ] 

PK.KO 

As  the  line  DL=LF,  it  is  obvious  that  ML  =  ^HF,  and  LN 
=  JDG;  therefore  ML.LN  =  iDG.HF.    Hence  IL  is  a  mean 
proportional  between  JDG  and  ^HF.    As  DL  a=  LF,  DL.LF 
s=  LF2.     Consequently,  the  above  analogy  is, 
As  L^2  :  LP  : :  DK.KF  :  KE2. 

Def.  6.  The  lines  LF,  LI,  are  called  the  first  and  second 
semi-axes ;  DK,  KF,  the  abscissas ;  and  KE  an  ordinate. 

The  property  of  the  ellipse,  demonstrated  in  this  article, 
therefore,  is  this : 

As  the  square  of  the  first  semi-axis  is  to  the  square  of  the 
second,  so  is  the  rectangle  of  the  two  abscissas  to  the  square 
of  the  ordinater 

It  is  observable  that  if  the  plane  DIEF  is  parallel  to  the 
base,  or  sub-contrarily  situated,  all  that  is  demonstrated  in 
this  article  continues  to  be  true  ;  but  in  either  of  those  cases 
the  curve  becomes  a  circle  (Arts.  80,  82) :  and  therefore  LI 
=  LF,  and  DK.KF  =  KE2. 


CONIC    SECTIONS. 


123 


Art.  85.  Let  ABC 
be  a  triangle,  formed 
by  the  section  of  a 
cone  and  a  plane  pass- 
ing through  its  axis, 
at  right  angles  to  the 
plane  of  its  base ;  and 
let  DFE  be  a  plane  at 
right  angles  to  the 
plane  of  the  triangle, 
so  situated  that  GF, 
the  common  section  of 
these  planes,  being 
produced,  will  meet 
CA,  the  opposite  side 
of  the  triangle  also 
produced,  beyond  the 
vertex  A ;  then  the 
curve  which  is  the 
common  section  of  the  conical  surface  and  the  plane  DFE,  is 
called  an  hyperbola ;  the  general  property  of  which  is  to  be 
shown. 

In  this  section,  take  any  point  N;  through  which  let  a 
plane  pass  parallel  to  the  plane  of  the  base ;  and  let  ILK  be 
the  common  section  of  this  plane  with  the  plane  of  the  tri- 
angle, and  NL  its  section  with  the  plane  DFE. 

Now  (18.2  sup.),  DG,  the  section  of  DFE  and  the  base  of 
the  cone,  and  NL,  are  both  at  right  angles  to  the  plane  of  the 
triangle.  Also  (Art.  80),  the  common  section  of  the  conical 
surface,  and  the  plane  which  passes  through  NL,  is  a  circle  ; 
consequently,  IL.LK^LN2,  and  BG.GC=GD2  (35.3).  Since 
IK  is  parallel  to  BC  (14.2  sup.),  by  similar  triangles, 

As  FG  :  FL  : :  BG  :  IL ; 


and 


As  HG  :  HL  : :  GC  :  LK ; 


124  SECTION   IV. 

consequently  (23.6), 
As  FG.HG  :  FL.HL  : :  BG.GC  :  IL.LK  : :  GD2  :  NIA 

Def.  7.  The  line  HG  is  called  the  axis  of  the  hyperbola ; 
HL,  FL,  as  likewise  HG,  FG,  corresponding  abscissas ;  and 
DG,  NL,  the  ordinates. 

The  property  of  the  hyperbola,  proved  in  this  article,  there- 
fore, is  this  :  The  rectangles  of  corresponding  abscissas  are 
to  each  other  as  the  squares  of  their  ordinates. 

From  what  has  been  demonstrated  in  the  last  six  articles, 
it  appears  that  tkere  are  five  different  figures  which  may  be 
formed  by  the  section  of  a  plane  and  the  surface  of  a  cone, 
viz.,  the  triangle,  circle,  parabola,  ellipse  and  hyperbola. 
The  properties  of  the  triangle  and  circle  being  explained  in 
common  Geometry,  the  remaining  three  are  usually  denomi- 
nated the  Conic  Sections.  A  few  of  the  most  useful  proper- 
ties of  these  figures,  deduced  from  the  general  relations 
above  demonstrated,  are  subjoined. 

Of  the  Parabola. 

Art.  86.  Let  BAC  be  a  parabola ;  AD,  part  of  the  axis,  an 

DC'2 
abscissa ;  DC  an  ordinate ;  AE  =  AF  ==  rrfv  EG  perpendi- 
cular, and  CG  parallel  to  ED ;  then,  FC  being  joined,  FC 
shall  be  equal  to  CG. 

Since  EF  is  bisected  in  A, 

4AF.AD  +  DF3  =  ED3  (8.2)  =  CG3. 
But  (47.1), 

FC3  =  DC3  +  DF3  =  (by  construction)  4AF.AD  +  DF3. 
Therefore,  CG3  =  CF* ;  and  CG  =  CF.  Q.  E.  D. 

Def.  8.  The  line  EG  is  called  the  directrix ;  4AE,  the  latus 
rectum;  and  the  point  F,  the  focus. 


CONIC    SECTIONS.  125 

E  I  P      G      R 


A 

TC  / 

r             X 

H 

X. 

\    \ 

M\ 

H^ 

3 

) 

c 

\J 

The  proposition  demonstrated  in  this  article,  therefore,  may 
be  enunciated :  If  from  any  point  in  the  parabola,  two  straight 
lines  be  drawn,  the  one  to  the  focus  and  the  other  at  right 
angles  to  the  directrix,  they  will  be  equal  to  each  other. 

Art.  87.  A  right  line  KFH,  drawn  through  the  focus 
parallel  to  the  directrix,  bounded  at  both  ends  by  the  para- 
bola, is  equal  to  the  latus  rectum;  and  the  rectangle  of  the 
latus  rectum  and  abscissa  is_equal  to  the  square  of  the  corre- 
sponding ordinate. 

It  is  evident  from  Art.  83,  that  KF  =  FH ;  and  (Art.  86), 
FH  =  HI  =  2AF  ;  therefore,  KH  =  4AE. 

Draw  any  other  ordinate  LM ;  then  (Art.  83), 

As  DC2  :  LM*  : :  AD  :  AL  ::  (1.6)  AD.KII  :  AL.KH. 

But  (by  construction  of  Art.  86), 

4AD.AF  =  AD.KH  =  DC2  .-.  AL.KH  =  LM2. 

Q.  E.  D. 

Art.  88.  If  a  point  be  taken  either  within  or  without  a 
parabola,  and  from  it  a  straight  line  be  drawn  to  the  focus, 
and  another  at  right  angles  to  the  directrix ;  the  former  of 

17 


126  SECTION    IV. 

these  lines  will  be  less  or  greater  than  the  latter,  according 
as  the  point  is  within  or  without  the  parabola. 

First,  let  N  be  taken  within  the  parabola ;  join  FN,  and 
produce  it  till  it  meets  the  parabola  in  C  ;  let  NP  and  CG  be 
at  right  angles  to  the  directrix ;  and  join  CP.  Then  CP  is 
greater  than  CG  (17.1  and  19.1) ;  but  CN-HNP>CP.  (20.1) 
>CG;andCG  =  CF  =  CN  +  NF.-.  CN  +  NP>CN  +  NP; 
and  therefore  NP>NF. 

Next,  let  O  be  without  the  parabola ;  then,  a  similar  con- 
struction being  used,  OG  >  OR ;  but  OC  +  CG  >  OG.  Also 
OC  +  CG  =  OC  +  CF  (Art.  86)  =  OF ;  therefore,  OF  >  OR. 

Q.  E.  D. 

Cor.  Hence,  a  point  is  either  in,  within  or  without  a  para- 
bola, according  as  the  line  drawn  from  it  to  the  focus  is  equal 
to,  less  or  greater  than  the  perpendicular  falling  from  it  upon 
the  directrix. 

Art.  89.  Let  D  be  a  point  in  the  parabola ;  DF  the  line  to 
the  focus  ;  DB  the  perpendicular  to  the  directrix ;  and  DG 
a  line  bisecting  the  angle  FDB ;  then  DG  touches  the  para- 
bola. 

In  DG,  take  any  other  point  I ;  and  join  IF,  IB ;  then  the 
angle  FDB  being  bisected,  we  have  (Art.  86)  the  sides  BD, 
DI,  and  the  contained  angle  BDI,  severally  equal  to  FD,DI, 
and  the  contained  angle  FDI;  consequently,  BI  =  FI  (4.1). 
But  BI  is  evidently  greater  than  IL,  the  perpendicular  from 
I  to  the  directrix  (19.1) ;  hence  the  point  I  is  without  the 
parabola  (Art.  88,  Cor.)  ;  and  therefore  DG  touches  the  para- 
bola. Q.  E.  D. 

Cor.  1.  A  right  line  through  the  vertex,  at  right  angles  to 
the  axis,  is  a  tangent  to  the  parabola.  For  AM  being  drawn 
through  the  vertex  A,  at  right  angles  to  EF,  it  is  evident  that 
every  point  in  AM,  except  the  point  A,  is  farther  from  F 
than  from  the  directrix. 


CONIC    SECTIONS. 
Or 


127 


L    B 


Cor.  2.  If  FB  be  drawn  from  the  focus  to  the  point  where 
the  line  through  D  parallel  to  the  axis  meets  the  directrix,  it 
is  manifest  that  FB  is  bisected  and  cut  at  right  angles  by  the 
tangent  DG.  For  the  triangles  BDH  and  FDH  are  in  every 
respect  equal  (4.1). 

Art.  90.  Let  DG  touch  the  parabola  in  D,  and  meet  the 
axis  in  G ;  then  DF  being  drawn  to  the  focus,  and  DN  at 
right  angles  to  the  axis,  FG  =  DF,  and  AG  =  AN. 

Because  DB  is  parallel  to  FG,  the  angle  FGD  =  BDG 
(29.1)  =  GDF  (Art.  89).     Hence  FG  "m  FD  (6.1). 

Again,  since  FD  =  DB  (Art.  86)  =  EN ;  and  EA  =  AF, 
AG  =  AN.  Q.  E.  D. 

Def.  9.  The  line  NG  is  called  the  subtangent ;  the  second 
part  of  this  article,  therefore,  shows  that  the  subtangent  is 
double  the  abscissa ;  or  GN  =  SAN. 

Def.  10.  The  line  DP,  drawn  from  the  curve  to  the  axis, 


128 


SECTION   IV. 


perpendicular  to  the  tangent,  is  called  the  normal ;  and  NP, 
the  segment  of  the  axis  between  the  ordinate  and  the  normal, 
is  called  the  subnormal. 

Art.  91.  The  subnormal  is  equal  to  half  the  latus  rectum. 

The  angle  GDP  being  a  right  angle,  is  equal  to  DGF  + 
DPF  (32.1) ;  and  GDF  =  DGF,  as  shown  in  the  last  article; 
the  remainder  FDP  =  DPF ;  .-.  FP  =  FD  (6.1)  fe  DB  (Art. 
86)  ==  NE.  Taking  FN  from  each,  NP  =  EF  =  £  the  latus 
rectum.  Q.  E.  D. 

Cor.  From  this  demonstration,  we  have  DF  =  FP. 

Abt.  92.  Let  DG  touch  the 
parabola  in  D,  and  FI  be  the 
perpendicular  from  the  focus  F 
to  the  tangent ;  A,  the  vertex ; 
then  shall  FI  be  a  mean  pro- 
portional between  DF  and 
FA. 

Join  AI,  and  draw  through 
D  the  ordinate  DN.  Then, 
since  the  angle  DIF  =  GIF ; 
and  DF  =  GF  (Art.  90) ;  DI 
=  IG.  Also,  AN  =  AG  con- 
sequently, AI  is  parallel  to  DN 
(2.6) ;  and  therefore  the  angles 
at  A  are  right  angles ;  where- 
fore the  triangle  IAF  is  similar  to  GIF  or  DIF  (8.6).    Hence, 

As  DF  :  FI  : :  FI  :  FA. 

Q.  E.  D. 
Cor.       As  DF  :  FA  : :  DF2  :  FI2  (cor.  2.20.6). 


Art.  93.  Let  UK  touch  the  parabola  in  H,  and  HR  be 
parallel  to  the  axis  AM ;  from  any  point  Q  in  the  parabola, 
let  QV  be  drawn  parallel  to  HK ;  then  QV,  produced,  will 


CONIC    SECTIONS. 


129 


meet  the  parabola  in  another  point ;  and  the  line  between  its 
points  of  section  with  the  curve  will  be  bisected  by  RH. 


From  Q,  draw  QS  at  right  angles  to  the  directrix ;  and 
from  the  centre  Q,  with  the  distance  QS,  describe  a  circle ; 
this  circle  will  evidently  pass  through  the  focus  F  (Art.  86), 
and  touch  the  directrix  in  S  (cor.  16.3).  Join  FR,  and  let 
QV  cut  FR  in  Y ;  then,  since  HK  is  at  right  angles  to  FR, 
and  also  bisects  it  (Art.  89,  cor.  2) ;  the  angles  at  Y  are  right 
angles ;  and  the  point  X,  where  the  circle  cuts  FY  the  second 
time,  lies  between  F  and  R.  Also,  FY  =  YX  (3.3).  Make 
RT  =  RS;  draw  TP  parallel  to  AM,  meeting  QV  in  P ;  and 
through  the  points  F,  T,  X,  describe  a  circle.  Then,  since 
FR  cuts  the  circle  FXS,  and  RS  touches  it,  FR.RX  =  RS2 
(36.3)  =  RT2 ;  consequently,  RT  touches  the  circle  FTX  in 
17 


130 


SECTION   IV. 


T  (37.3).  Hence  the  centre  of  that  circle  is  in  TP,  which  is 
at  right  angles  to  RT  (19.3) ;  it  is  also  in  VY,  which  bisects 
FX  at  right  angles  (cor.  1.3) ;  it  is  therefore  in  P;  the  point 
P  is  of  course  in  the  parabola  (Art.  86). 

Now,  PT,  HR  and  QS  being  parallel,  and  TR  =  RS,  it 
follows  (2.6)  that  PV  =  VQ.  Q.  E.  D. 

Def.  11.  Any  line  OR,  parallel  to  the  axis,  is  called  a  dia- 
meter of  the  parabola  ;  the  point  H,  where  the  diameter  meets 
the  curve,  is  called  its  vertex ;  4HR  is  called  the  latus  rectum 
of  that  diameter ;  the  line  PV  or  VQ,  parallel  to  the  tangent 
HK,  is  called  an  ordinate  and  HV  an  abscissa  to  the  diameter 
OH. 


Art.  94.  Let  PZ  be  drawn  at  right  angles  to  the  diameter 
OR,  and  PV  parallel  to  the  tangent  HK ;  then  PZ2=4AF.VH. 


CONIC    SECTIONS.  131 

Retaining  the  construction  of  the  last  article,  let  the  tan- 
gent  HK  meet  FR  in  a,  and  the  diameter  TP  in  c ;  draw  en 
parallel  to  FR,  and  join  Aa.  Then,  since  FR  is  bisected  in 
a  (Art.  89,  cor.  2) ;  and  FX  in  Y  (3.3) ;  Ya  or  nc  =  ^XR : 
also,  Pc  =  VH  (341).  But  (as  was  proved  in  Art.  92),  the 
triangle  FAa  is  similar  to  FaK  or  Yen.     Consequently, 

As  Fa  :  FA  : :  Pc  :  en ; 

whence  FA.Pc  (or  FA.VH)  =  Fa.nc  (16.6).     Hence, 

4FA.VH  =  FR.RX  =  TR2  =  PZ2 

Q.  E.  D. 

Art.  95.  If,  from  two  points  P,  A,  ordinates,  PV,  AO,  be 
drawn  to  any  diameter  OR,  the  squares  of  those  ordinates 
shall  be  to  each  other  in  the  same  ratio  as  their  abscissas  ; 
that  is, 

As  PV3  :  AO3  : :  VH  :  OH. 

Draw  PZ,  AW,  at  right  angles  to  OR ;  then  (Art.  94)  PZ3= 
4FA.VH;  and  AW*  =  4FA.OH;  consequently  (1.6), 

PZ3  :  AWa  : :  VH  :  OH. 

But  the  triangles  PVZ,  AOW,  being  similar, 

PV8  :  AO3  : :  PZ3  :  AW3  : :  VII  :  OH. 

Q.  K  D. 

Art.  96.  The  square  of  any  ordinate  is  equal  to  the  rec- 
tangle of  its  abscissa,  and  the  latus  rectum  of  the  diameter. 

Let  PV  be  an  ordinate  to  the  diameter  OH;  from  the 
vertex  of  the  axis  let  AO  be  drawn  to  the  same  diameter, 
parallel  to  PV  ;  from  H,  draw  HM  at  right  angles  to  the 
axis.     Then,  since  AOHK  is  a  parallelogram, 

OH  =  AK  -  (Art.  90)  AM  =  HW  .-.  OW  =  20H. 

Now, 


132  SECTION  IV. 

AO  -  AW3  +  OW2  (47.1)  =  4AF.0H  +  40H9  (Art.  94,  and 
cor.  2.8.2)  =  4RW.OH  +  4WH.OH  =  4RH.OH. 

But  (Art.  95), 

As  AO-  :  PV2  ::  OH  :  VH  ::  (1.6)  4RH.OH  :  4RH.VH. 

Hence,  PV2  =  4RH.VH.  Q.  E.  D. 


Art.  97.  A  double  ordinate  passing  through  the  focus  of  a 
parabola,  is  equal  to  the  latus  rectum  of  the  diameter  to 
which  that  ordinate  is  applied. 


Let  F  be  the  focus ;  VH  a  diameter ;  EG  the  directrix ; 
HD  a  tangent  to  the  parabola ;  PFO,  the  line  through  the 
focus  parallel  to  DH.  Join  FH ;  then,  PV  being  parallel  to 
DH,  the  angle  DHG  =  FVH;  and  DHF  =*  HFV  (29.1). 
But  DHG  =  DHF  (Art.  89) ;  therefore,  FVH  =  HFV;  and 
HV  =  FH  (6.1)  -  HG  (Art.  86).  Now,  PV  =  VO  (Art.  93), 
and  PV2  =  4GH.HV  (Art.  96)  =  4HG2.  Hence  PV  =  2HG 
(cor.  2.8.2).     Therefore,  PO  =  4HG. 

The  case  of  the  double  ordinate  applied  to  the  axis,  is 
proved  in  Art.  87.  Q.  E.  D. 


CONIC    SECTIONS. 


133 


H^^ 

Xi 

N 

F 

E 

M 

G  j 

X 

1> 


Of  the  Ellipse. 

Def.  12.  Let  ACBD  be  an 
ellipse;  AE,  CE,  the  semi- 
axes  ;  and  from  C,  the  extre- 
mity of  the  less,  with  a  dis- 
tance equal  to  the  greater, 
let  an  arc  be  described  cut- 
ting AB  in  F  and  G ;  each 
of  those  points,  F,  G,  is 
called  the  focus  of  the  ellipse;  and  the  line  HI,  passing 
through  either  focus  at  right  angles  to  AE,  meeting  the 
ellipse  in  H  and  I,  is  called  the  latus  rectum  of  the  ellipse. 

Art.  98.  The  rectangle  of  the  abscissas  AF.FB,  into  which 
the  axis  AB  is  divided  by  the  focus,  is  equal  to  the  square  of 
the  semi-axis  CE. 

Since  AB  is  bisected  in  E,  and  divided  unequally  in  F ; 

AF.FB  +  FE2  =  AE2  (5.2)  =  FC2  =  FE2  +  CE2  (47.1) ; 
therefore,  AF.FB  =  EC2.  Q.  E.  D. 

Cor.  Hence,  HF  is  a  third  proportional  to  AE,  EC.  For 
(Art.  84), 

As  AE2  :  EC2  ::  AF.FB  (EC2)  :  FH2; 
whence  AE  :  EC  : :  EC  :  FH. 


Art.  99.  If  from  L,  any  point  in  the  ellipse,  a  line  LN  be 
drawn  at  right  angles  to  the  second  or  minor  axis  CD ;  then, 

As  EC2  :  EB2  : :  CN.ND  :  NL2. 

Draw  LM  at  right  angles  to  AB;  then  (Art.  84), 

As  EB2  :  EC2  : :  AM.MB  (EB2  —  EM2)  :  ML2  or  EN2. 

Therefore  (19.5), 

As  EB2  :  EC2  : :  EM2  :  EC2  —  EN2  or  CN.ND  (5.2) ; 

18 


1U 


SECTION   IV. 


hence,  by  inversion, 

As  EC2  :  EB2  : :  CN.ND  :  EM2  or  NL2. 

It  is  therefore  manifest  that  the  property  demonstrated  in 
Art.  84  is  equally  true,  whichever  axis  is  divided  by  an 
ordinate. 

Cor.  Since  CN.ND  is  less  than  CE2  (27.6),  it  follows  that 
NL2  is  always  less  than  EB2,  and  consequently  NL  less  than 
EB. 

Art.  100.  If,  on  either  axis  of  an  ellipse,  a  circle  be 
described;  and  from  any  point  in  the  ellipse  a  perpendi- 
cular be  drawn  to  that 
axis,  meeting  the  circle  ;  it 
will  be,  as  the  axis  on 
which  the  circle  is  de- 
scribed is  to  the  other 
i 

axis,  so  is  the  ordinate  to 
the  circle,  to  the  ordinate 
to  the  ellipse. 

Let  ACBD  be  the  ellipse, 
and  first  let  the  circle  be 
described  on  the  greater  axis  AB,  and  let  LM  be  the  ordi- 
nate ;  then  (Art.  84), 

As  EB2  :  EC2  : :  AM.MB  :  ML2. 
But  AM.MB  =  MP2  (35.3).     Hence  (22.6), 
As  EB  :  EC  : :  MP  :  ML. 

Again,  let  the  circle  be  described  on  CD,  and  the  ordinate 
LN  meet  the  circle  in  I ;  then  (Art.  99), 

As  CE2  :  EB2  : :  CN.ND  :  NL2. 

But  CN.ND  =  NI2  (35.3) ;  therefore  (22.6), 

CE  :  EB  : :  NI  :  NL. 

Now,  CE  :  EB  : :  CD  :  AB. 

Hence  the  proposition  is  manifest. 


CONIC   SECTIONS.  135 

Cor.  Two  ordinates,  ML,  TR,  being  drawn  to  the  same 
axis,  ML  :  TR  : :  MP  :  TS. 

Art.  101.  The  sum  of  the  lines  FH,  GH,  drawn  from  any 
point  in  the  ellipse  to  the  two  foci,  is  equal  to  the  greatei 
axis  AB. 

Take  E  the  middle  of 
AB,*  and  through  it  draw 
the  perpendicular  CD ;  this 
will  be  the  less  axis.  From 
H  draw  the  ordinate  HI, 
and  take  EL  a  fourth  pro- 
portional to  EB,  EG  and  EI. 
Then  (22.6), 

As  EB2:  EG2::  EI2:  EL2. 

Consequently  (19.5), 

EB2 :  EG2  : :  EB2  — EI2 :  EG2  — EL2  (that  is,  5.2)  : :  AI.IB 
:  FL.LG. 

Hence  (17.5  and  Def.  12), 

EB2  :  EC2  ::  AI.IB  :  AI.IB  — FL.LG. 

Therefore  (Art.  84), 

AI.IB  —  FL.LG  =  IH2. 

Again,      BE2  f  EL2  =  2BE.EL  +  BL2  (7.2) ; 

also,  GE2  +  EI2  =  2GE.EI  +  GI2. 

Taking  the  latter  of  these  equations  from  the  former,  and 
remembering  that  BE2  —  EI2=  AI.IB  (5.2) ;  that  GE2  — EL2 
=  FL.LG ;  and  that  BE.EL  ==  GE.EI  (16.6) ;  we  have 

AI.IB  —  FL.LG  =  BL2  —  IG2 ; 

that  is,  IH2  =  BL2  —  IG2.  Hence,  IH2  +  IG2  =  BL2,  Con- 
sequently  (47.1),  GH2  =  BL2;  and  GH  =  BL. 


*  The  middle  of  the  axis  is  usually  called  the  centre  of  the  ellipse. 


136 


SECTION    IV. 


Taking  AE2  +  EL2  +  2AE.EL  =  AL2  (4.2),  and  FE2  + 
EP  +  2FE.EI  =  FI2,  and  proceeding  as  before,  we  have 

IH2  =  AL2  —  FP  .-.  FP  +  IH2  =  AL2; 

whence  (47.1),  FH2  =  AL2;   therefore,  FH  =  AL.     Conse- 
quently, 

FH  +  GH  =  AL  +  BL  =  AB. 

Q.  E.  D. 

Cor.  1.  Since  FH  =  AL,and  GH  =  BL,  it  is  manifest  that 

EL  =  FH  —  AE  =  AE  —  GH. 

Hence,  FH  —  GH  ==±  2EL ;  or  EL  =  $(FH  —  GH). 

Cor.  2.  If,  from  a  point  without  the  ellipse,  two  right  lines 
be  drawn  to  the  foci,  their  sum  will  be  greater  than  the 
greater  axis  of  the  ellipse ;  but  if  a  point  be  taken  within  the 
ellipse,  the  sum  of  the  lines  drawn  from  it  to  the  foci  will  be 
less  than  the  greater  axis.  This  is  evident,  from  what  is 
above  proved  and  21.1. 

Cor.  3.  Conversely  :  A  point  is  either  in,  without  or  within 
an  ellipse,  according  as  the  sum  of  the  lines  drawn  from  it  to 
the  foci  is  equal  to,  greater,  or  less  than  the  greater  axis. 


Art.  102.  If,  from  any  point  P  of  an  ellipse,  a  straight  line 
PR  =  AE,  half  the  greater  axis,  be  applied  to  the  less  axis 

K  JT      CD,  cutting  the  greater  axis 

in  S ;  then  shall  SP  =  EC, 
half  the  less  axis. 

From  P  draw  PT  at  right 
angles  to  AB ;  then,  because 
of  the  similar  triangles  TSP, 
ESR: 


As  PR:  PS  ::  ET:ST; 
hence  (22.6  and  19.5), 


CONIC   SECTIONS.  •  137 

As  PR2(  =  AE2)  :  PS2  ::  AE2  —  ET2  :  PS2—  ST2  ::  (5.2 

and  47.1)  AT.TB  :  TP2  ::  (Art.  84)  AE2  :  EC2; 
therefore,  PS  =  EC.  Q.  E.  D. 

The  ellipsograph,  or  instrument  for  describing  an  ellipse, 
is  founded  upon  this  property. 

Art.  103.  From  a  point  H  in  an  ellipse,  two  lines  HF, 
HG,  being  drawn  to  the  foci,  and  one  of  them  FH  produced, 
the  line  HV,  which  bisects  the  exterior  angle,  is  a  tangent  to 
the  ellipse. 

Make  HW,  in  FH  produced,  =  HG ;  join  GW,  cutting  the 
bisecting  line  in  V ;  in  VH,  take  any  point  K ;  and  join  KG, 
KW  and  KF.  Then  the  triangles  GHV,  WHV,  have  GH= 
WH,  and  the  angle  GHV=WHV;  hence  (4.1),  GV=WV, 
and  the  angle  GVH  =  WVH.  Consequently,  GK=WK; 
and,  therefore, 

FK  +  GK  =  FK  +  WK. 
But  FK  +WK>FW  (20.1),  and  FW=FH+HG=AB  (Art. 
101);  therefore  FK  +  GK>AB;  and  (Cor.  3,  Art.  101)  the  point 
K  is  without  the  ellipse.     This  being  true  of  every  point  in 
HV,  except  H,  the  line  HV  is  a  tangent  to  the  ellipse. 

Q.  E.  D. 

Cor.  1.  From  this  demonstration  it  is  obvious  that  the  line 
HV,  bisecting  the  exterior  angle  GHW,  also  bisects  GW,  and 
cuts  it  at  right  angles  in  V. 

Cor.  2.  Hence  the  angles  FHK,  GHV,  which  the  lines  from 
the  foci  to  the  point  of  contact  make  with  the  tangent,  are 
equal.     For  WHV  =  KHF  (15.1). 

Art.  104.  Aright  line  through  the  vertex  of  either  axis, 
parallel  to  the  other  axis,  is  a  tangent  to  the  ellipse. 

The  line  DM  parallel  to  AB,  makes  the  angle  GDM=DGF, 
and  MDN  =  GFD  (29.1),  and  DGF  =  GFD  (5.1) ;  conse- 
quently, DM  bisects  the  angle  GDN,  and  is  therefore  a  tan- 
gent to  the  ellipse  (Art.  103). 

18  at* 


138 


SECTION   IV. 


Next,  take  BY  parallel  to  CD,  and  through  any  point  X 
in  CD  draw  XZ  parallel  to  BE,  meeting  the  ellipse  in  Z ; 
then  XZ  is  less  than  EB,  (Cor.,  Art.  99);  that  is,  Z  lies  between 
DC  and  BY ;  therefore,  the  line  BY  is  without  the  ellipse. 


Art.  105.  Let  the  right 
lines  FW  and  GW,  of  which 
FW=AB,  the  greater  axis, 
be  drawn  from  the  foci  to 
meet  in  W ;  and  let  VH,  bi- 
secting GW  at  right  angles 
in  V,  cut  FW  in  H ;  then 
HV  touches  the  ellipse  in 
H.  Join  HG ;  then  it  is  evi- 
dent (4.1)  that  GH  =WH, 
and  the  angle  GHV=WHV;  hence  the  angle  GHW  is  bisect- 
ed by  the  line  HV  ;  and  since  HG  =  HW, 

FH  +  HG  =  FW  =  AB; 

therefore  (Art.  101,  Cor.  3),  the  point  H  is  in  the  ellipse; 
and  (Art.  103)  HV  is  a  tangent  to  the  ellipse. 

Cor.  A  line  which  cuts  GW  at  right  angles,  but  does  not 
bisect  it,  is  not  a  tangent  to  the  ellipse.  For  this  line  is 
parallel  to  HV,  and,  if  it  cuts  GW  between  V  and  W,  it  does 
not  meet  the  ellipse ;  but  if  it  falls  between  G  and  V,  it  must 
cut  the  ellipse. 

Art.  106.  A  straight  line  which  meets  an  ellipse,  but  does 
not  bisect  the  exterior  angle  formed  by  the  lines  drawn  to 
the  foci,  is  not  a  tangent  to  the  ellipse. 

Let  HR  meet  the  ellipse  in  H,  but  not  bisect  the  exterior 
angle  GHW.  On  HR  let  fall  the  perpendicular  GN;  and 
produce  it  to  M,  making  NM  =  GN;  join  FM,  meeting  HR 
in  I ;  join  also  GI*  GH  and  HM.  Then  (4.1),  GI  =  IM,  GH 
=HM,  and  the  angle  GHN=MHN.     Hence  the  angle  GHM 


CONIC    SECTIONS. 


139 


is  bisected  by  the  line  HR,  but  by  supposition  the  angle 
GHW  is  not ;  consequently,  HM  does  not  coincide  with 
HW ;  and  therefore  FHM  is  a  triangle,  of  which  FM  is  less 
than  FH  +  HM  (20.1).     But 

FH  +  HM  =  FH  +  HG  =  AB  (Art.  101). 

Also  FM  =  FI  +  IG.  Therefore,  FI  +  IG  are  less  than  AB, 
and  the  point  I  is  within  the  ellipse  (Art.  101,  Cor.  3).  Hence 
HR  is  not  a  tangent. 

Cor.  From  this  and  Art.  103,  it  is  evident  that  the  tangent 
must  bisect  the  exterior  angle  formed  at  the  point  of  contact 
by  right  lines  drawn  to  the  foci ;  and  that  only  one  right  line 
can  touch  the  ellipse  at  a  given  point. 


Art.  107.  Let  HM  touch  the  ellipse  in  H,  and  meet  the 
greater  axis  AB  produced  in  M ;  and  from  H  let  HI  be  drawn 
at  right  angles  to  AB ;  then,  E  being  the  centre,  EI,  EB  and 
EM  shall  be  proportionals. 

Take  EL  a  fourth  proportional  to  EB,  EG  and  EI ;  this 
line  is  half  the  difference  between  FH  and  GH  (Art.  101, 


140 


SECTION   IV. 


Cor.  1).  Now,  the  tangent  HM  bisects  the  exterior  angle  of 
the  triangle  FHG  (Cor.  2,  Art.  103),  and, meets  the  base  pro- 
duced; therefore  (A.  6), 

AsFH  :  HG  ::  FM  :  MG; 

consequently  (E.  5  and  Art.  101), 

As  2EB  :  2EL  ::  FM+MG  (2EM)  :  FG  (2EG) ; 

wherefore  (15.5), 

AsEB  :  EL  ::  EM  :  EG? 

consequently  (1.6), 

As  EB3  :  EB.EL  : :  EM.EI  :  EG.EI. 

But  EB.EL  =  EG.EI  (16.6) ;  therefore,  EB3  =  EM.EI ;  and 

(17.6),  EI  :  EB  : :  EB  :  EM. 

Q.  E.  D. 

Cor.  If  the  tangent  MH  meet  the  less  axis  EC  produced 
in  P,  and  HN  be  drawn  at  right  angles  to  EC ;  then  shall 
EN,  EC  and  EP  be  proportionals. 


CONIC    SECTIONS.  141 

By  similar  triangles, 

As  PE  :  PN  ::  EM  :  NH  or  EI  ::  (1.6)  EM.EI  :  EI2  :: 

(by  this  article)  EB2  :  EP. 
Hence  (D.  5), 
As  PE  :  EN  : :  EB3  :  EB3  — EP  : :  (5.2,  Art.  84  and  16.5) 

EC3  :  EN2. 

Again  <1.6),     As  PE  :  EN  : :  PE.EN  :  EN2. 
Consequently,  PE.EN  =  EC*;  or  EN  :  EC  : :  EC  :  EP. 

Art.  108.  Let  MHP  touch  the  ellipse  in  H,  and  meet  the 
axes  in  M  and  P ;  then  if,  on  the  axes,  circles  be  described, 
cutting  the  ordin^tes  HI  and  HN  in  R  and  O,  the  lines  MR 
and  PO  shall  touch  the  circles. 

Join  ER ;  then,  since  ER  =  EB,  we  have  (Art.  107), 

As  EI  :  ER  : :  ER  :  EM. 

Hence  (6.6),  the  triangles  EIR,  ERM,  are  similar;  conse- 
quently, ERM  is  a  right  angle ;  wherefore  (cor.  16.3),  RM 
touches  the  circle.  In  the  same  manner  it  may  be  proved 
that  OP  touches  the  circle. 


Art.  109.  Let  PT  be  a  right  line  touching  the  ellipse  in 
O ;  FP,  GT,  perpendiculars  falling  upon  it  from  the  foci  F 
and  G ;  then  shall  FP.GT  =  EC2,  the  square  of  the  less  semi- 
axis. 

Join  FO,  and  let  FO, 
GT,  produced,  meet  in  W; 
and  join  OG,  ET.  Then, 
since  the  angle  GOT  = 
WOT;andGTO=WTO; 
the  side  OG  =  OW,  and 
GT=WT(26.1).  Hence 
FW  =  FO  +  OG  =  (Art. 
101)  AB.  Now,  in  the 
triangles  FGW,  EGT,  FG 


142 


SECTION    IV. 


=  2EG,  GW  ==  2GT,  and  the  angle  at  G  common ;  therefore 
(6.6),  FW  =  2ET,  or  EB  =  ET.  Consequently,  the  circle 
described  on  AB  will  pass  through  T.  In  the  same  manner 
it  may  be  proved  that  it  will  pass  through  P. 

Next,  produce  TE  till  it  meets  the  circle  in  S,  and  join 
FS ;  then,  because  ES  =  ET,  EF  =  EG,  and  the  angle  SEF 
==  TEG  (15.1) ;  the  side  FS  must  =  GT,  and  the  angle  EFS 
=  EGT  (4.1);  consequently,  FS  is  parallel  to  GT  (27.1)- 
that  is,  FS  is  in  the  same  straight  line  with  PF.  For  GT 
and  FP  are  at  right  angles  to  the  same  line,  and  are  therefore 
parallel. 


Now,    FP.FS  ==  AF.FB  (35.3)  =  EC2  (Art.  98) ; 


therefore,  FP.GT  ==  EC2. 


Q.  E.  D. 


Art.  110.  Let  RK  be  a  tangent  to  the  ellipse  in  H;  the 
li-ne  HN  a  perpendicular  to  RK,  meeting  the  greater  axis  in 
M,  and  the  less  in  N;  HI,  HP,  ordinates  to  the  axes  AB,  CD ; 
then  it  will  be, 

AsEB2  :  EC2  : :  EI  :  IM; 


and 


As  EC2  :  EB2  : :  EP  :  PN. 


Join  FH,  GH ;  and  pro- 
duce FH  to  L,  making  HL 
=  HG ;  join  GL,  and  draw 
EO  parallel  to  HM.  Then 
GL  is  at  right  angles  to 
HK  (Cor.,  Art.  103) ;  and 
therefore  parallel  to  HM  or 
OE.  Hence  EO,  which  bi- 
sects FG,  also  bisects  FL 
(2.6);  wherefore  FO=EB. 
Then,  HL  being  =  HG,  and  OL  =  FO;  HO  =  half  the  dif- 
ference of  FH  and  HG.     Consequently  (Art.  32), 

AsEB  :  EG  ::  EI  :  OH; 


CONIC    SECTIONS.  143 

also  (2.6), 

As  FO  (EB)  :  FE  (EG)  : :  OH  :  EM; 

hence  (23.6), 

AsEB2  :  EG2  ::  EI  :  EM; 

therefore  (D.  5), 

As  EB2  :  EB2— EG2  : :  EI  :  IM. 
But  EB2  —  EG2  =  EC2  (see  def.  12).     Consequently, 
As  EB2  :  EC2  : :  EI  :  IM. 

Again,  fr.om  similar  triangles, 

As  EI  (PH)  :  IM  ::  NP  :  HI  (PE); 
and  by  inversion, 

IM  :  EI  : :  PE  :  PN. 
Hence,  EC2  :  EB2  : :  PE  :  PN.  Q.  E.  D. 


c        7 

— -yjH\ 

M 

II 

in 

i 

aI   ] 

f- 

E 

Xi  . 

I    Or 

B 

Art.   111.    Let 

FK  be  the  ordinate 
through  the  focus 
F;  KR  a  tangent 
to  the  ellipse  at  K 
cutting  the  greater 
axis  AB  produced 
in  R;  arnd  RM 
a  perpendicular  to 
AB ;  then,  if  from  any  point  H  in  the  ellipse,  HF  be  drawn 
to  the  focus,  and  HM  at  right  angles  to  RM,  it  will  be, 

As  EA  :  EF  : :  HM  :  HF  : :  AR  :  AF. 

From  H  draw  HI,  at  right  angles  to  AB ;  and  take  EL  a 
fourth  proportional  to  AE,  FE  and  EI.  Then  (Art.  101), 
FH  =  AL ;  also  (Art.  107), 

As  ER  :  EA  : :  EA  :  EF  : :  (by  construction)  EI  :  EL ; 


M4  SECTION  IV. 

therefore  (12.5), 

As  ER  :  EA  : :  ER  +  EI  :  EA  +  EL. 

But  ER  +  EI  =  HM ;  and  EA  +  EL  =  FH.     From  the 
analogy  ER  :  EA  : :  EA  :  EF,  we  have  (19.5), 

ER  :  EA  : :  AR  :  AF. 

Hence,        EA  :  EF  : :  HM  :  HF  : :  AR  :  AF. 

Q.  E.  D. 

Def.  13.  If  we 
make  EN  =  ER, 
and  draw  NP  pa- 
rallel to  RM,  each 
of  the  lines  RM, 
NP,  is  called  a  di- 
rectrix to  the  el- 
lipse; and  if  HP, 
HG,  are  drawn,  the 

latter  to  the  focus  And  the  former  at  right  angles  to  the 

directrix,  it  may  be  proved  as  before  that 

HP  :  HG  : :  BN  :  BG. 
Cor.  Because  ER  :  EA  : :  EA  :  EF; 
(16.6)  EA2  =  ER.EF  =  EF2  +  EF.FR ; 

therefore,  AE2  — EF2  =  EF.FR;  that  is,  EC2  =  EF.FR. 


M 

y<             C 

1                       i 

T 

U; 

ii 

AV  3 

p             E 

X.  . 

t    Cr  J'JBj 

i) 


Art.  112.  Retaining  the  construction  of  the  last  article, 
it  will  be 


FH  = 


EC2 


EG2 


AE  —  EF.cos  BFH    t    AE  +  EF.cos  AFH 
From  Art.  28, 

As  1  :  cos  BFH  : :  FH  :  FI  =  FH.cos  BFH. 
Now, 

MH  |  RF  +  FI  =  RF  +  FH.cos  BFH; 


CONIC    SECTIONS.  143 

therefore  (Art.  Ill), 

As  AE  :  FE  : :  RF  +  FH.cos  BFH  :  FII ; 
consequently  (16.6), 

FH.AE  i  RF.FE  +  FH.FE.cos  BFH; 
whence, 
FH.AE— FH.FE.cos  BFH=RF.FE=(Cor.,  Art.  Ill)  EC2. 

Consequently, 

EC2 EC2       

FH  -  AE  —  FE.cos  BFH  "  AE  +  FE.cos  AFH 

For,  cos  AFH  *  —cos  BFH.  Q.  E.  D 

Cor.  If  IH  be  produced  to  meet  the  semicircle  on  AB  in 
T,  and  ET  be  joined,  then  shall 

FH  =  AE  +  EF.cos  BET  =  AE  — EF.cos  AET. 
By  Art.  28, 
As  1  :  cos  BET  ::  ET  (=AE) :  EI  : :  EF  :  EL=EF.cos  BET. 
But 
FH=AL  (Art.  101)=AE  +  EF.cos  BET=AE— EF.cos  AET. 

Art.  113.  Every  diameter*  to  an  ellipse  is  bisected  in  the 
centre. 

Let  HI,  passing  through  the  centre  E,  meet  the  ellipse  in 
H  and  I ;  then  is  EI  =  EH.  For  if  it  is  not,  take  EP=EH  5 
and  from  H,  I,  P,  draw  lines  to  the  foci  F  and  G.  Then  we 
shall  have  EH  =  EP ;  EF  =  EG ;  and  the  angle  HEF  = 
PEG  (15.1) ;  therefore  (4.1),  FH  =  PG.  In  like  manner,  GH 
==  FP ;  therefore, 

FH  +  HG  =  FP  +  PG. 

But,  FH  +  HG  =  FI  +  IG  (Art.  101) ; 

consequently,  FP  +  PG  =  FI  +  IG ; 

which  is  absurd  (21.1). 

*  Any  right  line  passing-  through  the  centre,  limited  at  both  extremities 
by  the  ellipse,  is  called  a  diameter. 

19  N 


143 


SECTION   IV. 


Art.  H4.  The  tangents  to  an  ellipse  passing  through  the 
extremities  of  any  diameter,  are  parallel  to  each  other. 

Let  HM  and  IN  touch  the  ellipse  in  H  and  I,  the  extremi- 
ties of  the  diameter  HI ;  and  produce  FH,  GI,  the  lines  from 
the  foci,  to  K  and  L ;  then  the  angles  GHK,  FIL  are  bisect- 
ed by  the  tangents  HM,  IN  (Cor.  1,  Art.  106).  And  since 
GE,  EH  are  respectively  equal  to  FE,EI  (Art.  112),  and  the 
angle  GEH  =  FEI  (15.1),  the  angle  EHG  =  EIF,  and  the 
side  HG  =  FI  (4.1).  In  the  same  manner,  FHE  =  GIE; 
consequently,  the  whole  angle  FHG  =  FIG ;  and,  therefore, 
(13.1)  GHK  =  FIL;  whence  GHM  *a  FIN.  But  EHG  = 
EIF;  therefore,  EHM  =  EIN;  whence  (27.1)  HM  is  parallel 
to  IN. 

Cor.  Hence,  if  tangents  be  drawn  through  the  extremities 
of  any  two  diameters,  they  will  form  a  parallelogram. 

Def.  14.  If  the  diameter  OT  be  drawn  parallel  to  the  tan- 
gents through  the  extremities  of  IH,  then  OT  is  said  to  be 
conjugate  to  IH. 

Art.  115.   Let  OT,  which  is   conjugate  to  IH,  cut  the 

radius  vector,  or  line  from  the  focus  to  the  curve,  FH  in  V; 
then  is  HV  =  AE,  half  the  greater  axis. 

Through  the  other  focus  G,  draw  GW  parallel  to  OT,  cut- 
ting FH  in  W.  Then,  because  OT,  and  consequently  WG, 
is  parallel  to  the  tangent  HM,  the  angle  HWG  =  KHM ; 


CONIC    SECTIONS. 


147 


and  HGW^GHM  (29.1) ;  therefore  (Cor.  %  Art.  103),  HWG 
=HGW;  and  consequently  HG  =  HW  (6.1).  Since  EV  is 
parallel  to  GW,  and  FE  =  EG ;  therefore  (2.6),  FV  =  VW. 
Hence,  VH=FV+HG;  and  therefore  (Art.  101)  VH=AE. 

Q.  E.  D. 

Art.  116.  If  the  diameter  OT  is  parallel  to  HM,  the  tan- 
gent at  H,  thei#the  diameter  HI  shall  be  parallel  to  TN,  the 
tangent  at  T. 


s 

Q 

^-"~c 

//$) 

\^ 

/     K 

Ii  \ 

JN      ^^ 

^\\T 

^c 

D 

t 

11 

Let  HM,  TN,  meet  the  greater  axis  AB  produced  in  M 
and  N ;  through  H  and  T  draw  the  ordinates  HK,  TL,  to 
that  axis ;  and  let  them  meet  the  circle  described  on  AB,  in 
Q  and  P;  join  QM,  QE,  PN  and  PE ;  then  QM  and  PN  are 
tangents  to  the  circle  (Art.  108).  Now,  ET  being  parallel 
to  HM,  the  angle  TEL  «  HMK  (29.1),  and  ELT  -  MKH; 
therefore  (4.6), 

As  EL  :  LT  ::  MK  :  KH; 

and,  alternately, 

EL  :  MK  : :  LT  :  KH  : :  (Cor.,  Art.  100)  LP  :  KQ. 

Hence,  As  EL  :  LP  : :  MK  :  KQ; 

and  the  angles  at  L  and  K  are  equal ;  therefore  (6.6),  the 
angle  LEP  =  KMQ ;  consequently,  EP  is  parallel  to  QM 


148 


SECTION   IV. 


(27.1).  But  EQM  is  a  right  angle  (18.3) ;  therefore,  QEP  is 
also  a  right  angle  (29.1) ;  and  EPN  is  a  right  angle  (18.3) ; 
therefore,  PN  is  parallel  to  EQ  (28.1).  Consequently,  the 
angle  LNP  =  KEQ  (29.1) ;  and  as  NLP  =  EKQ,  we  have, 

As  LN  :  EK  : :  LP  :  QK  : :  (Cor.,  Art.  100)  LT  :  KH. 

Hence,  LN  :  LT  : :  EK  :  KH; 

consequently,  the  triangles  LNT,  KEH  are  similar  (6.6),  and 
the  angle  LNT  =  KEH;  therefore  (27.1),  TN  is  parallel  to 
EH.  Q.E.D. 

Cor.  1.  Hence,  OT  being  conjugate  to  IH,  IH  is  also  con- 
jugate to  OT. 

Cor.  2.  Hence,  also,  if  through  the  extremities  H  and  T 
of  two  conjugate  diameters,  ordihates,  HK,  TL,  be  drawn  to 
the  greater  axis,  meeting  the  circle  described  on  that  axis  in 
Q  and  P;  the  tangent  QM  to  the  circle  is  parallel  to  the 
radius  EP,  and  the  tangent  PN  to  the  radius  EQ. 

Art.  117.  The  sum  of  the  squares  of  any  two  semi-conju- 
gate diameters,  is  equal  to  the  sum  of  the  squares  of  the 
semi-axes. 


t""li" 

Q 

//£> 

^S 

/      K 

Ii  \ 

IU      "^^ 

/x 

X>w 

\r 

D 

___^^y/v 

K 

m 


Let  OT,  IH,  be  conjugate  diameters ;  then,  retaining  the 
construction  of  the  last  article,  the  triangle  EMH  is  similar 


CONIC    SECTIONS.  149 

to  NET,  because  EH  is  parallel  to  TN,  and  HM  to  ET. 
The  triangle  HMK  is  likewise  similar  to  TEL.  Conse- 
quently, 

As  EM  :  MH  ::  EN  :  ET; 

and  AsMH  :  MK  ::  ET  :  EL; 

therefore  (22.5), 

As  EM  :  MK  : :  EN  :  EL  : :  (Art.  107,  and  Cor.  2  to  20.6) 
EB2  :  EL2. 

Also,  As  EM  :  EK  : :  EB2  :  EK2. 

Consequently  (24.5), 

EM  :  MK  +  EK  : :  EB2  :  EL2  +  EK2; 

wherefore,  EB2  =  EL2  +  EK2. 

If  the  tangents  MH,  NT,  produced,  meet  the  other  axis 
CD  produced,  in  S  and  R ;  and  the  ordinates  HW  and  TX 
be  drawn  to  that  axis ;  we  have,  in  like  manner,  the  triangles 
ESH,  SHW,  respectively  similar  to  RET  and  ETX ;  whence, 
as  before, 

AsES  :  SW  ::  ED2  :  EX2; 

and  As  ES  :  EW  : :  EC2  :  EW2. 

Consequently, 

EC2  =  EX2  +  EW2  =  LT2  +  KH2. ' 

Wherefore, 

EB2  +  EC2  -  EL2  +  LT2  +  EK2  -f  KH2  =  (47.1)  ET2+EH2. 

Q.  E.  D. 

Cor.  Since  EK2+EL2=EB2=EQ2=EP2,  it  follows  (47.1) 
that  EL  ==  KQ,  and  EK  =  LP.     Hence, 

As  EB  :  EC  : :  EL  :  KH  : :  EK  :  TL. 

In  like  manner, 

As  EC  :  EB  : :  EX  :  WH  : :  EW  :  XT. 
20 


150 


SECTION    IV. 


Art.  118.  The  parallelogram  formed  by  the  tangents 
through  the  extremities  of  any  two  conjugate  diameters,  is 
equal  to  the  rectangle  of  the  axes. 


/    c 

^<? 

0/ 

r       s 

^T 

Q 


D 


Let  ET,  EH  be  two  conjugate  semidiameters ;  TM,  HM, 
tangents  to  the  ellipse  passing  through  their  extremities ;  Q, 
the  intersection  of  HM  with  the  greater  axis  produced. 
Draw  HS,  TR  at  right  angles  to  AB,  the  greater  axis ;  and 
EX  at  right  angles  to  HM.     Then  (Cor.,  Art.  117), 

EC  :  EB  ::  TR  :  ES; 

and  (Art.  107),      EB  :  EQ  : :  ES  :  EB ; 

therefore  (22.5),     EC  :  EQ  : :  TR  :  EB. 

But  in  the  similar  triangles  EQX,  TER ; 

AsEQ  :  EX  ::  ET  :  TR; 

therefore  (23.5),    EC  :  EX  : :  ET  :  EB; 

consequently  (16.6), 

EB.EC  ==  ET.EX  =  the  parallelogram  ETMH. 

Now,  this  parallelogram  being  one-fourth  of  that  which  cir- 
cumscribes the  ellipse,  and  EB.EC  one-fourth  of  the  rectan- 
gle of  the  axis,  it  is  obvious  that  the  circumscribing  paral- 
lelogram is  equal  to  the  rectangle  AB.CD. 


CONIC   SECTIONS. 


151 


n 


/  c 

o// 

\X 

H 

jfy/'k 

N^N 

1^- — 

p\ 

12 


Art.  119.  Let  OT,  IH  be  conjugate  diameters;  MN,  a 
tangent  at  H ;  FH,  GH,  lines  from  the  foci  to  the  point  of 

contact ;  then  shall  the  rect- 
angle FH.HG-ET2.  Draw 
FM,  GN,  HP  and  EX  at 
right  angles  to  MN  or  OT 
and  let  OT  cut  FH  in  W 
thenisWH=AE(Art.H5) 
and  OT  being  parallel  to 
MN,  the  angle  HWE  = 
WHM  (29.1).  ButWHM 
=  GHN  (Cor.  2,  Art.  103); 
consequently,  the  triangles  FHM,  GHN  and  HWP,  being 
right  angled  at  M,  N  and  P,  are  similar;  therefore  (4.6), 

As  HP  (or  EX)  :  HW  (or  EB)  : :  FM  :  FH  : :  GN  :  GH; 

consequently  (22.6), 

As  EX2 :  EB2  : :  FM.GN  :  FH.GH  ::  (Art.  109)  EC* :  FH.GH. 

But  (Art.  118  and  22.6), 

As  EX2  :  EB2  ::  EC2  :  ET2; 

therefore,  FH.GH  -  ET2. 

Cor.  From  the  similar  triangles  FHM,  GHN ;  we  have, 
FH  :  GH  : :  FM  :  GN  : :  (1.6)  FMe  :  FM.GN  or  EC2. 

Art.  120.  Let  OT,  IH  be  conjugate  diameters;  HP,  a 
perpendicular  to  OT,  cutting  the  greater  axis  in  V;  then 
shall  HP.HV  ==  EC2,  the  square  of  the  less  semi-axis. 

Let  the  tangent  at  H  meet  the  less  axis  produced  in  S ; 
and  draw  HK,  HL  at  right  angles  to  the  axes ;  and  EX 
parallel  to  PH.  Then  the  line  ES  being  parallel  to  KH,  and 
EX  to  VH,  the  angle  SEX  =  VHK  (29.1),  rfnd  the  angles  at 
X  and  K  are  right  ones  ;  therefore, 

SE  :  EX  (or  PH)  : :  HV  :  HK  or  EL. 


152  SECTION  IV. 

Therefore, 

PH.HV  =  SE.EL  =  EC2  (Cor.,  Art.  107). 

Art.  121.  Let  CD,  FG  be  conjugate  diameters;  DL,  a 
tangent  to  the  ellipse ;  HI,  a  chord  parallel  to  DL,  cutting  CD 
in  K ;  then  shall  HI  be  bisected  in  K ;  and 

As  DE2  :  EG2  : :  DK.KC  :  HK2. 


Let  DL,  HI  meet  the  greater  axis  AB  produced  in  L  and 
V;  through  H,  D,  I  and  G,  draw  HM,  DO,  IN  and  GW  at 
right  angles  to  AB,  meeting  the  circle  on  AB,  in  R,  P,  S  and 
W;  join  PL,  RS,  SV,  EP  and  EW;  and  let  RS  cut  EP  in 
T ;  and  join  TK.  Now,  since  IV  is  parallel  to  DL,  the  angle 
NVI  =  OLD  (29.1),  and  VNI  =  LOD,  both  being  right  an- 
gles; therefore  (4.6), 

As  VN  :  NI  ::  LO  :  OD. 

But  (Cor.,  Art.  100), 

AsNI  :  NS  ::  OD  :  OP; 
therefore  (22.5), 

As  VN  :  NS  ::  LO  :  OP; 

consequently  (6.6),  the  angle  NVS  =  OLP.     Hence  (28.1), 
VS  is  parallel  to  LP. 


CONIC    SECTIONS.  153 

Again,  from  similar  triangles, 

As  VM  :  MH  ::  VN  :  NI; 

and  (Cor.,  Art.  100), 

MH  :  MR  ::  NI  :  NS; 

therefore  (22.5), 

As  VM  :  MR  : :  VN  :  NS. 

If,  therefore,  we  suppose  VR  joined,  the  angle  MVR  =  NVS 
(6.6) ;  consequently,  VS  and  SR  are  in  the  same  straight  line. 
Now,  as  DL  touches  the  ellipse  in  D,  PL  touches  the  circle 
in  P  (Art.  108).  Hence,  EPL  is  a  right  angle  (18.3) ;  and 
RS,  which  is  parallel  to  PL,  is  bisected  in  T  (3.3).  Since 
HV  is  parallel  to  DL,  and  TV  to  PL ;  we  have  (2.6), 

As  EK  :  KD  ::  EV  i  VL  : :  ET  :  TP; 

hence  TK  is  parallel  to  PD  (2.6),  and  consequently  to  SI  and 
RH;  wherefore, 

ST  :  TR  : :  IK  :  KH. 

But  ST  =  TR;  therefore,  IK  =  KH. 

Again,  since  EW  is  parallel  to  PL  (Cor.  2,  Art.  116),  and 
therefore  to  RV ;  and  EG  to  DL  or  HV ;  the  angle  EVR  = 
VEW,  and  EVH  =  VEG ;  hence,  HVR  =  GEW.  Also,  HR 
being  parallel  to  GW,  the  angle  VHR  —  EGW ;  for  each  of 
them  is  equal  to  GtiR  (29.1).     Hence, 

EW  :  EG  : :  VR  :  VH  : :  RT  :  HK. 

Now,  from  similar  triangles, 

As  ED  :  EP  ::  EK  :  ET; 

and,  therefore  (22.6), 

As  ED2  :  EP2  : :  EK2  :  ET2  : :  (19.5) 

ED2  — EK2  :  EP2  — ET2  ::  (5.2  and  47.1)  DK.KC  :  RT2. 

But  EW2  or  EP2  :  EG2  : :  RT2  :  HK2 : 

20 


154  SECTION   IV. 

therefore  (22.5), 

As  ED2  :  EG2  : :  DK.KC  :  HE 


Q.  E.  D. 


Of  the  Hyperbola. 

Art.  122.  Let  ABC  be  a  triangle 
formed  by  the  section  of  a  cone  and 
a  plane  passing  through  its  axis,  at 
right  angles  to  the  plane  of  its  base ; 
and  let  a  plane  at  right  angles  to 
the  plane  of  the  triangle  cut  the 
cone  in  DFE  and  the  opposite 
cone,  made  by  the  extension  of  the 
sides  of  the  given  cone  on  the  other 
side  of  A,  in  OHR ;  then  (Art.  85) 
the  curves  DFE  and  OHR  form 
opposite  hyperbolas. 

The  two  branches  of  either  hy- 
perbola, as  likewise  the  two  opposite 
hyperbolas,  are    like   figures,   and 
equal  to  each  other. 
Let  BDC  be  the  base  of  the  cone ;  DE,  the  intersection  of 
BDC  and  the  plane  DFE ;  and  GFHP,  the  intersection  of  the 
cutting  plane  and  the  plane  of  ABC. 

In  GF  take  any  point  L,  make  HP  =  FL,  and  through  L 
and  P  let  planes  pass  parallel  to  the  base  of  the  cone ;  then 
the  sections  KNI,  MRT,  of  these  planes  and  the  conical  sur- 
face, are  circles,  whose  centres  are  in  KI,  MT  (Art.  80),  the 
intersections  of  these  planes  with  the  plane  of  ABC,  which 
passes  through  the  axis  of  the  cone.  But  (18.2  sup.)  SLN 
and  OPR  are  at  right  angles  to  the  plane  of  ABC,  and  there- 
fore (Def.  1.2  sup.)  to  KI  and  MT  in  that  plane;  therefore 
(3.3),  SL  =  LN,  and  OP  =  PR.  Hence,  the  two  branches 
of  either  hyperbola  are  equal  to  each  other. 


CONIC   SECTIONS.  155 

Again,  since  the  planes  MOT  and  KSI  are  parallel,  MT  is 
parallel  to  KI  (14.2  sup).  Hence  the  triangles  FLI  and  MFP 
are  similar,  as  likewise  the  triangles  KHL  and  PHT ;  there- 
fore, 

AsFL  :  FP  ::  LI  :  MP; 

and  AsLH  :  HP  ::  KL  :  PT; 

consequently, 

As  FL.LH  :  FP.HP  : :  LI.KL  :  MP.PT  : :  (35.3)  LN2 :  PR2. 

But,  by  construction,  FL  ===  HP ;  therefore,  LH  ±=  FP,  and 
FL.LH  =  FP.HP;  .-.  LN2  =  PR2,  or  LN  =  PR.  Conse- 
quently, equal  ordinates  corresponding  to  equal  abscissas,  the 
opposite  hyperbolas  must  be  like  figures  and  equal  to  each 
other.  Q.  E.  D. 

Art.  123.  Let  MAN,  PBQ  (see  fig.  on  page  156)  be  two 
hyperbolas,  formed  by  the  sections  of  a  cone  and  a  plane,  as 
described  in  Art.  122 ;  AB,  the  interval  between  the  points 
where  the  cutting  plane  cuts  the  opposite  cones,  correspond- 
ing to  HF  in  the  figure  on  page  154. 

Bisect  AB  in  E ;  draw  DEC  at  right  angles  to  AB ;  and, 
drawing  the  ordinate  HI  from  any  point  H  in  the  curve,  to 
meet  BA  produced  in  I,  make  ED  and  EC  such  that 

AI.IB  :  IH2  ::  EA2  :  ED2  or  EC2; 

then  AB  is  called  the  first,  and  DC  the  second,  axis  of  the 
hyperbola.  The  points  A,  B,  are  called  the  vertices,  and  E 
the  cenfre,  of  the  hyperbola. 

From  this  construction  and  Art.  85,  it  is  obvious  that  AE2 
is  to  ED2  as  the  rectangle  of  any  corresponding  abscissas  is 
to  the  square  of  their  ordinate. 

Art.  124.  Def.  15.  Join  AD ;  and  make  EF,  EG,  each 
equal  to  AD ;  then  F  and  G  are  called  the  foci  of  the  hyper- 


156 


SECTION    IV. 


M. 


bolas;   and  the  double  ordinate  RFS,  passing  through  the 
focus,  is  called  the  latus  rectum. 

The  latus  rectum  is  a  third  proportional  to  the  first  and 
second  axes.     For  BA  is  bisected  in  E ;  therefore, 

BF.FA  +  AE2  =  EF2  (6.2)  =  AD2  =  AE2  *  ED2  (47.1) ; 

consequently,  BF.FA  =  ED2;  hence  (Art.  123), 

AE2  :  ED2  ::  ED2  :  RF2  or  FS2; 

wherefore  (22.6), 

AE  :  ED  ::  ED  :  RF  or  FS; 
and  AB  :  DC  : :  DC  :  RS. 

Cor.  From  this  demonstration,  it  is  obvious  that 
ED2  =  AF.FB  =  BG.GA. 


CONIC    SECTIONS.  157 

Art.  125.  If  from  any  point  H  in  the  hyperbola,  a  perpen- 
dicular HK  be  let  fall  on  the  second  axis,  it  will  be, 

As  ED2  :  AE2  : :  ED2  +  EK2  :  HK2. 

For  (Art.  123), 

As  AE2  :  ED2  : :  AI.IB  :  IH2  : :  (12.5) 
AE2  +  AI.IB  :  ED2  +  IH2  or  ED2  +  EK2. 

But  (6.2),  AE2  +  AI.IB  =  EI2. 

Hence,  by  inversion, 

ED2  :  AE2  : :  ED2  +  EK2  :  EI2  or  HK2. 

Q.  E.  D. 

Art.  126.  The  difference  of  two  right  lines,  HG,  HF, 
drawn  from  any  point.  H  in  the  hyperbola  to  the  foci,  is  equal 
to  AB,  the  first  axis. 

Take  EL  a  fourth  proportional  to  EA,  EF  and  Ei ;  HI 
being  an  ordinate  to  the  point  H.     Then  (22.6), 

AsEA2  :  EF2  ::  EI2  :  EL2; 
therefore  (17.5), 
AsEA2  :  EF2  — EA2  (ED2)  ::  EI2  :  EL2— EI2  ::  (19.5) 
EI2  —  E A2  :  EL2  —  EI2  —  ED2. 

But  (6.2),  EI2  —  EA»  =  ALBI; 

and  (Art.  123), 

AsEA2  :  ED2  ::  AI.IB  :  III2; 

therefore,  EL2  —  EF  —  ED2  =*  IH2. 

Again,      EA2  +  EL2  ==  2AE.EL  +  AL2  (7.2)  ; 

and  EF2  +  EI2  =  2EF.EI  +  FI2. 

Taking  the  difference  of  these  equations,  and  remembering 
that  EF2  —  EA2  =  ED2;  and  AE.EL  =  EF.EI,  because 
AE  :  EF  : :  EI  :  EL ;  we  have, 

EL2  —  EI2  —  ED2  =  AL2  —  FI2. 
21 


158 


SECTION    IV. 


Hence,  AL2  —  FP=IH2 ;  or  AL2  =  IH2  +  FP  =  (47.1)  FH2 ; 
or  AL  =  FH. 

Further,  EL2  +  EB2  +  2BE.EL  =  BL2  (4.2) ; 
also,  EI2  +  EG2  +  2IE.EG  £  IG2. 

Taking  the  difference  as  before, 

EL2  —  EI2  —  ED2  =  BL2  —  IG2. 
But  EL2  —  EI2  —  ED2  =  IH2 ; 

therefore,  BL2  — IG2=IH2; 

and,  therefore, 

BL2  =  IG2  +  IH2  =  (47.1)  GH2;  or  BL  =  GH. 
Consequently, 

BA=(BL  —  AL=)(>H  —  FH. 

Q.  E.  D. 


CONIC   SECTIONS.  159 

Art.  127.  If,  from  a  point  Z  within  an  hyperbola,  two 
right  lines,  FZ,  GZ,  be  drawn  to  the  foci,  the  difference  of 
these  lines  is  greater  than  AB,  the  first  axis ;  but  if  from  a 
point  b  without  the  hyperbola,  two  right  lines,  bG,  bF,  be 
drawn  to  the  foci,  the  difference  of  these  lines  will  be  less 
than  the  axis  AB. 

First,  let  ZG  meet  the  hyperbola  in  a,  and  join  Fa ;  then, 
since  FZ  is  less  than  Fa  +  aZ,  the  difference  between  GZ 
and  FZ  is  greater  than  between  GZ  and  Fa  +  aZ ;  that  is, 
than  Ga  —  Fa.  But  (Art.  126)  Ga  —  Fa  =  AB;  therefore, 
GZ  —  FZ  is  greater  than  AB. 

Next,  let  Fb  meet  the  hyperbola  in  d,  and  suppose  Gd 
joined;  then,  because  Gd  is  greater  than  Gb  —  bd,  Gd  —  dF; 
that  is,  AB  is  greater  than  Gb  —  bF,  or  Gb  —  bF  is  less  than 
AB.  Q.  E.  D. 

Cor.  Hence  a  point  is  either  in,  within,  or  without  an  hy- 
perbola, according  as  the  difference  of  the  lines  drawn  from 
it  to  the  foci  is  equal  to,  greater,  or  less  than  the  first  axis. 

Art.  128.  If,  from  any  point  H  (see  fig.  on  page  160)  in 
the  hyperbola,  a  right  line  HM  be  drawn  bisecting  the  angle 
FHG,  made  by  Ikies  to  the  foci  F,  G,  the  line  HM  will  be  a 
tangent  to  the  hyperbola. 

Take  on  HG,  the  line  HL  =5  HF ;  and  take  in  MH  any 
other  point  P,  and  join  PL,  PF ;  then  (4.1),  FP=LP.  Now, 
since  HF  ==  HL,  LG  must  be  equal  to  AB  (Art.  126) ;  hence, 

AB  =  PL  +  LG  —  FP. 
But  PL  +  LG  are  greater  than  the  right  line  joining  P  and 
G ;  hence  the  excess  of  that  line  above  PF  is  less  than  AB ; 
consequently,  the  point  P  is  without  the  hyperbola  (Cor.,  Art. 
127).  And  this  being  true  of  every  point  in  PM  except  the 
point  H,  that  line  must  be  a  tangent  to  the  hyperbola. 

Q.  E.  D. 

Cor.  1.  From  this  we  may  infer  that  the  tangent  must 
bisect  the  angle  FHG ;  for  if  it  was  possible  to  draw  a  tan- 


160 


SECTION  IV. 


gent  through  H  which  did  not  bisect  the  angle,  we  might 
have  two  right  lines  touching  the  same  curve  in  the  same 
point,  and  yet  not  coinciding  with  each  other. 

Cor.  2.  The  line  ka  through  the  vertex  of  the  hyperbola, 
at  right  angles  to  GF,  is  a  tangent ;  for  the  angles  GAa,  FAa, 
are  equal. 

Art.  129.  Let  FN,  GI  be  perpendiculars  falling  from  the 
foci  F,  G,  upon  a  tangent  HI ;  then  shall  FN.GI  =  ED2,  the 
square  of  the  second  semi-axis. 

Take,  as  in  the  last  article,  HL  ==  HF ;  join  LN,  NE ;  and 
produce  NE  to  meet  GI  in  K.  Then,  since  HF  =  HL ;  and 
the  angle  FHN  =  LHN  (Cor.  1,  Art.  128) ;  the  angle  HNF 
must  be  equal  to  HNL  (4.1) :  consequently,  HNL  =  a  right 
angle,  and  therefore  FNL  is  a  right  line  (14.1).  Now,  in  the 
triangles  NFE,  LFG,  we  have  the  angle  at  F  common,  and 
the  sides  NF,  FE,  the  halves  of  LF,  FG,  respectively; 
whence  (6.6)  the  angle  FNE  =  FLG,  and  NE  =  half  LG; 


CONIC   SECTIONS.  1G1 

consequently  (28.1),  NK  is  parallel  to  LG.  But  LN  is  pa- 
rallel to  GK;  hence  (34.1),  LN  =  GK,  and  NK  =  LG  =  AB 
(Art.  126).     Hence, 

EN  =  EK  =  JAB  =  EA. 

Consequently,  a  circle  described  from  the  centre  E,  at  the 
distance  EA,  will  pass  through  N  and  K ;  it  will  also  pass 
through  I,  because  KIN  is  a  right  angle  (converse  of  31.3). 
Therefore  (cor.  36.3),  AG.GB  =  IG.GK;  that  is,  (Cor.,  Art. 
124),  ED2  =  FN.GL  Q.  E.  D. 

Art.  130.  Let  HM  touch  the  hyperbola  in  H,  and  meet 
the  first  axis  AB  in  M,  and  HS  be  an  ordinate  to  that  axis ; 
then  it  shall  be, 

As  ES  :  EA  :  :  EA  :  EM. 

Take  ER  a  fourth  proportional  to  EA,  EF  and  ES ;  then,  as 
proved  in  Art.  126,  AR  =  FH,  and  BR  =  GH.  Then,  the 
vertical  angle  of  the  triangle  GHF  being  bisected  by  the  line 
HM  (Cor.  1,  Art.  128), 

AsGH  :  HF  ::  GM  :  MF  (3.6) ; 

hence  (E.  5) 

GH+  HF  :  GH  — HF  ::  GM  +  MF  :  GM  — MF; 

then,  taking  the  halves  of  these  quantities, 

ER  :  EA  ::  EF  :  EM; 

and,  alternately  (16.5), 

ER  :  EF  : :  EA  :  EM. 

But  ER  :  EF  ::  ES  :  EA; 

therefore  (11.5),    ES  :  EA  : :  EA  :  EM.  Q.  E.  D. 

Art.  131.  Let  FO,  Go,   be  the  ordinates  through  the  foci ; 
OT,  ot,  tangents  to  the  hyperbolas  at  O  and  o,  cutting  the 
first  axis  in  T  and  t ;  QTU  and  qtu,  perpendiculars  to  AB ; 
21  o* 


102 


SECTION    IV. 


then,  taking  any  point  H  in  the  hyperbola,  and  drawing  HQ</ 
parallel  to  the  axis  AB,  it  will  be, 

As  FH  :  HQ  ::  FA  :  AT  : :  GH  :  Bq  ::  GA  :  BT. 

Draw  HS  at  right  angles  to  the  axis;  and  take  ER  a 
fourth  proportional  to  EA,  EF,  ES ;  then  (Art.  126)  AR  = 
FH,  and  BR  =  GH.     Now,  by  construction, 

ER  :  ES  : :  EF  :  EA  : :  (Art.  130)  EA  :  ET  : :  (19.5) 
ER  — EA  :  ES  — ET  ::  EF— EA  :  EA  — ET; 
for  the  last  four  terms  substituting  their  equals, 

FH  :  HQ  ::  AF  :  AT. 
Again  (12.5), 

ER  :  ES  ::  ER+EA  :  ES+ET  ::  EF+EA  :  EA+ET. 
Hence,  GH  :  Uq  : :  FB    (GA):  BT. 

And  these  ratios  in  both  cases  are  the  same  as  EF  :  EA. 


CONIC   SECTIONS. 


163 


Each  of  the  lines  QU  and  qu  is  called  the  directrix  of  the 
hyperbola. 

This  article  being  compared  with  articles  86  and  111,  it  is 
manifest  that  from  any  point  in  either  of  the  three  conic 
sections,  two  straight  lines  being  drawn,  one  of  them  to  the 
focus  and  the  other  at  right  angles  to  the  directrix,  they  will 
have  to  each  other  a  constant  ratio.  In  the  parabola,  the 
perpendicular  upon  the  directrix  is  equal  to,  in  the  ellipsis 
it  is  greater,  and  in  the  hyperbola  it  is  less,  than  the  radius, 
vector,  or  line  front  the  focus  to  the  curve. 


Art.  132.  Let  AB,  DC  be  the  axes  of  the  hyperbolas;  AH, 
AI,  at  right  angles  to  AB,  and  each  equal  to  half  DC ;  then, 
right  lines  being  drawn  through  E,  the  middle  of  AB,  and 
the  points  H  and  I,  and  indefinitely  extended,  they  are  called 
the  asymptotes ;  the  property  of  which,  to  be  demonstrated 
hereafter,  is,  that  they  continually  approach  the  hyperbola, 
but  do  not  meet  it. 

The  asymptotes  do  not  meet  the  hyperbola. 


164  SECTION   IV. 

From  any  point  N  in  the  hyperbola,  draw  NM  at  right 
angles  to  the  axis,  and  let  it  meet  the  asymptote  in  X ;  then 
(from  similar  triangles  and  22.6), 

As  EA2  :  AH*  : :  EM2  :  MK2. 

But  (Art.  123), 

As  EA2  :  ED2  (AH2)  ::  AM.MB  :  MN2; 

and  EM2  is  greater  than  AM.MB  (6.2) ;  therefore,  MK2  is 
greater  than  MN2,  and  MK  greater  than  MN. 

Art.  133.  Retaining  the  construction  in  the  last  article, 
produce  KM  to  meet  the  asymptote  EQ  in  L;  then  shall 
KN.NL  =  ED2. 

As  in  last  article,  we  have, 

EA2  :  ED2  : :  EM2  :  MK2  : :  AM.MB  :  MN2  : :  (19.5) 
EM2— AM.MB  :  MK2— MN2 : :  (6.2  and  5.2)  AE2  .  KN.NL. 
Hence,  KN.NL  =  ED2  =  HA.AI. 

Cor.  Hence,  OPQ  being  drawn  parallel  to  KL,  the  rect- 
angle OP.PQ  =  KN.NL;  and,  therefore  (16.6), 

As  KN  :  OP  : :  PQ  :  NL. 

Art.  134.  The  asymptotes  continually  approach  to  the 
hyperbola. 

Taking  KL  and  OQ  as  in  the  last  article,  it  is  evident  that, 
ER  being  greater  than  EM,  PQ  is  greater  than  NL ;  but, 

KN  :  OP  ::  PQ  :  NL; 

hence,  OP  is  less  than  KN. 

Art.  135.  Through  the  vertex  A,  and  any  other  point  N 
of  the  hyperbola,  let  the  lines  AS,  NT  be  drawn  parallel 
to  one  of  the  asymptotes  EQ,  meeting  the  other  in  S  and 
T;  then. 

As  ES  :  ET  : :  TN  :  SA. 


CONIC   SECTIONS. 


165 


Draw  AW,  NV  parallel  to  EO;  then  the  triangles  IAW, 
LNV,  are  similar ;  as  are  also  HAS,  KNT ;  hence  the  follow- 
ing analogies : 

As  AW  :  NV  : :  AI  :  NL  : :  (Art.  133)  KN  :  HA  : :  TN  :  AS. 

Therefore,  As  ES  :  ET  : :  TN  :  SA. 

Cor.  If  PU  be  drawn  parallel  to  EQ,  then 

EU  :  ET  : :  TN  :  UP. 


Scholium.  From  the  property  demonstrated  in  Art.  135  is 
deduced  a  relation  between  logarithms  and  the  areas  con- 
tained between  the  hyperbolic  curve  and  its  asymptote.  Let 
EA  =s  ED,  and  consequently  ES  =  SA,  and  SEW  a  right 
angle.  The  hyperbola  is  then  called  an  equilateral  or  rect- 
angular one.  If  in  that  case  we  assume  ES  =  1 ;  and  of 
course  the  square  SW  also  =  1 ;  then  ET  being  estimated  in 
units  of  ES,  and  the  area  ASTN  in  units  of  SW,  it  is  proved 
by  writers  on  differentials  that  ASTN  is  the  logarithm  of 

22 


166  SECTION    IV. 

ET,  provided  the  modulus  (Art.  15)  m  =  1.     Hence,  those 
logarithms  are  termed  hyperbolic. 

It  is,  however,  observable,  that  these  hyperbolic  areas  may 
be  made  to  express  logarithms  of  other  kinds.  For,  if  the 
relation  between  the  axes  is  such  that,  ES  being  =  1,  the 
area  of  the  parallelogram  SW  shall  be  expressed  by  the 
modulus,  the  area  of  ASTN  will  be  the  logarithm  of  ET, 
according  to  the  system  to  which  that  value  of  m  belongs. 
But  the  demonstration  of  these  properties  would  lead  further 
into  the  differential  and  integral  calculus,  than  the  design  of 
this  work  admits. 


SECTION  V. 


SPHERICAL  PROJECTIONS. 

Article  136.  The  business  of  Spherical  Projections  is  to 
represent  by  lines,  drawn  or  described  on  a  plane  given  in 
position,  the  circles  which  are  described  on  the  surface  of  a 
sphere.  The  lines  thus  drawn  or  described  on  the  plane,  are 
called  the  projections  of  the  circles  which  they  represent ; 
and  are  so  framed  that,  to  an  eye  properly  located,  every 
circle  on  the  sphere  will  appear  coincident  with  its  repre- 
sentative. 

Def  1.  The  plane  on  which  the  circles  of  the  sphere  are 
represented,  is  called  the  plane  of  projection ;  and  the  point 
where  the  eye  is  supposed  to  be  located,  the  projecting  point. 
A  right  line  drawn  from  the  projecting  point  to  any  point  on 
the  sphere,  and  extended  to  meet  the  plane  of  projection,  is 
termed  a  projecting  line. 

Def.  2.  Every  circle  on  the  sphere  is  called  an  original 
circle;  and  the  figure  which  represents  it  on  the  plane  of 
projection,  a  projected  circle. 

Def.  3.  In  the  orthographic  and  stereographic  projections, 
the  plane  of  projection  is  supposed  to  pass  through  the  centre 
of  the  sphere.  Then  the  common  section  of  this  plane  and  the 
spherical  surface  is  a  circle,  wThich  is  called  the  primitive 
circle.  This  circle  is  evidently  a  great  one  (Art.  45) ;  and, 
being  both  on  the  sphere  and  plane  of  projection,  may  be 
considered  as  an  original  circle,  projected  into  itself. 

Def.  4.  In  the  orthographic  projection,  the  projecting  point 
is  in  the  axis  of  the  primitive  circle ;  but  so  remote,  that  all  the 

(167) 


168 


SECTION   V. 


projecting  lines  drawn  to  the  different  points  of  the  sphere 
may  be  considered  as  parallel. 

Def  5.  In  the  stereographic  projection,  the  projecting  point 
is  at  one  of  the  poles  of  the  primitive  circle. 

Def.  6.  The  line  of  measures  of  any  circle  which  is  to  be  pro- 
jected, is  the  common  section  of  the  plane  of  projection,  and 
another  plane  which  passes  through  the  axes,  both  of  the 
primitive  circle  and  of  the  circle  to  be  projected. 

Def.  7.  The  semitangent  of  an  arc  is  the  tangent  of  half  the 
arc ;  not  half  the  tangent  of  the  arc. 


Of  the  Orthographic  Projection. 

Art.  137.  If  a  right  line  AB  be  projected  orthographically 
upon  a  plane,  the  projection  will  be  a  right  line ;  and  the 
original  line  will  be  to  its  projection,  as  radius  to  the  cosine 
of  the  inclination  of  the  original  to  the  plane  of  projection. 

Let  EF  be  the  plane 
of  projection,  seen 
edgewise ;  Aa,  Bb, 
the  projecting  lines, 
through  the   extremi- 

ties   of   AB,   meeting 

the  plane  of  projection 
in  a,  b.  Conceive  a  plane  to  pass  through  Aa,  Bb ;  this  plane 
will  include  AB,  and  be  at  right  angles  to  the  plane  of  pro- 
jection (def.  4  and  17.2  sup).  The  common  section  of  these 
planes  will  evidently  be  the  projection  of  AB ;  but  this  sec- 
tion is  a  straight  line  (3.2  sup.)  contained  between  Aa  and 
Bb ;  that  is,  it  is  the  line  ab. 

Through  A  draw  AD  parallel  to  ab;  then  is  DAB  the 
inclination  of  AB  to  the  plane  of  projection.     And  (Art.  28) 

radius  :  cosine  DAB  : :  AB  :  AD  or  ab. 


E 


SPHERICAL    PROJECTIONS. 


169 


Cor.  1.  When  a  line  is  parallel  to  the  plane  of  projection, 
its  projection  is  equal  to  the  original  line. 

Cor.  2.  When  two  lines  which  make  an  angle  with  each 
other  are  parallel  to  the  plane  of  projection,  their  projections 
make  an  equal  angle  with  each  other.  This  is  obvious  from 
9.2  sup. 

Cor.  3.  Any  figure  which  is  delineated  on  a  plane  parallel 
to  the  plane  of  projection,  is  projected  into  a  figure  similar 
and  equal  to  itself. 

Art.  ,138.  A  circle  whose  plane  is  at  right  angles  to  the 
plane  of  projection,  is  projected  into  a  right  line  equal  to  its 
diameter. 

Let  EF  be  the 
plane  of  projec- 
tion, seen  edgewise ; 
ABDC,  the  circle; 
Gil,  IK,  right  lines 
perpendicular  to  the 
plane  of  projection, 

•^     ^*^-: ^     ^  touching  the  circle 

in  A  and  D ;  then  the  plane  GHKI  will  evidently  include  the 
circle ;  and  its  intersection  HK  with  the  plane  of  projection, 
must  be  the  projection  of  the  circle :  but  HK  =  AD,  the 
diameter  of  the  circle. 

Cor.  Hence  any  figure  which  is  delineated  on  a  plane  at 
right  angles  to  the  plane  of  projection,  is  projected  into  a 
right  line. 

Art.  139.  A  circle  of  the  sphere,  whose  plane  is  parallel 
to  the  plane  of  projection,  is  projected  into  a  circle  equal 
to  itself,  and  concentric  with  the  primitive  circle. 

For  every  circle  of  the  sphere  which  is  parallel  to  the 
plane  of  projection,  has  the  same  axis  with  the  primitive 
circle ;  and  that  axis  being  at  right  angles  to  the  plane  of 
22  p 


Or 

B 

T 

A 

(            \ 

D 

V       / 

170 


SECTION   V. 


projection,  is  a  projecting  line  (def.  4).  Consequently,  the 
centre  of  the  original  circle  must  be  projected  into  the  centre 
of  the  primitive.  And  every  radius  of  that  circle  is  projected 
into  a  line  equal  to  itself  (Cor.  1,  Art.  137). 

Cor.  As  the  radius  of  any  circle  on  the  sphere  is  the  sine 
of  its  distance  from  its  own  pole ;  the  radius  of  a  projected 
circle  whose  original  is  parallel  to  the  primitive,  is  the  sine 
of  the  distance  of  that  original  circle  from  its  pole. 

Art.  140.  A  circle  whose  plane  is  inclined  to  the  plane  of 
the  primitive,  is  projected  into  an  ellipse,  whose  major  axis 
is  equal  to  the  diameter  of  the  original  circle,  and  whose 
minor  axis  is  to  the  major,  as  the  cosine  of  the  inclination  of 
the  planes  is  to  radius. 

Let  AGBH  be  the  ori- 
ginal circle ;  P,  its  centre ; 
GH,  its  diameter  parallel 
to  the  plane  of  projection ; 
AB,  the  diameter  at  right 
angles  to  GH ;  ABba,  a 
plane  at  right  angles  both 
to  the  plane  of  projection 


- — if. "" 

g 

and  to  the  plane  of  AGBH.  From  any  point  D  in  the  origi- 
nal circle,  let  DE  be  drawn  perpendicular,  and  DQ  parallel 
to  BA ;  and  let  the  given  circle  be  projected  into  the  figure 


agbh;  the  lines  Aa,  Gg,  Bfr,  Wi,  P/?,  Q?,  Del,  Ee,  being  the 
projecting  lines,  which  (def.  4)  are  necessarily  at  right  angles 
to  the  plane  of  projection.  Then,  since  GH  and  DE  are 
parallel  to  the  plane  of  projection,  their  projections  are  equal 
to  the  lines  themselves  (Cor.  1,  Art.  137) ;  that  is,  gp  =  GP; 
ph  =  PH ;  de  =  (qp=)  DE  =  QP.     Hence, 

gq.qh  =  GQ.QH  =  (35.3)  DQ2. 

But  Pp,  Ee,  and  Bb,  in  the  plane  aABb,  being  parallel, 

BP  :  EP  ::  bp  :  ep  (2.6) ; 


SPHERICAL   PROJECTIONS.  171 

consequently  (22.6), 

BP2  :  EF  (or  DQ3)  ::  bp2  :  ef  or  dq2; 

that  is,  gp2  :  gq.qh  : :  bp2  :  dq2 ; 

wherefore  agbh  is  an  ellipse,  whose  major  axis  is  gh  =  GH, 
the  diameter  of  the  original  circle  (Art.  84).  Also,  gh 
(  as  AB)  :  ab  : :  radius  :  cosine  of  the  inclination  of  AB  to 
the  plane  of  projection  (Art.  137).  Now,  the  plane  AGBH 
and  the  plane  of  projection  are  both  at  right  angles  to  the 
plane  aABb ;  hence,  if  those  planes  were  extended  so  as  to 
meet,  their  common  section  would  be  at  right  angles  to  the 
same  plane  (18.2  sup.),  and  therefore  at  right  angles  to  AB. 
Hence,  the  inclination  of  AB  to  the  plane  of  projection  is  the 
inclination  of  the  original  circle  AGBH  to  the  same  plane. 

Cor.  1.  The  major  axis  of  the  ellipse  into  which  a  circle  is 
projected,  is  twice  the  sine  of  the  arc  of  a  great  circle  inter- 
cepted between  the  original  circle  and  its  own  pole. 

Cor.  2.  The  minor  axis  of  the  same  ellipse,  or  that  axis 
produced,  passes  through  the  centre  of  the  primitive  circle. 
For  the  plane  ABba,  being  perpendicular  to  the  circle  AGBH, 
and  passing  through  its  centre,  must  pass  through  its  axis ; 
and  that  axis  passes  through  the  centre  of  the  primitive 
circle. 

Cor.  3.  The  distances  of  the  extremities  of  the  minor  axis 
from  the  centre  of  the  primitive  circle,  are  the  sines  of  the 
greatest  and  least  distance  of  the  original  circle  from  the 
pole  of  the  primitive. 


SECTION   V. 


Of  the  Stereographic  Projection. 

Art.  141.  To  explain  the  nature  of  this  projection,  let 
ABED  be  a  circle  formed  by  the  section  of  a  spherical  sur- 
face and  a  plane  which 
passes  through  the  centre 
of  the  sphere;  this  plane 
is  here  represented  by  the 
plane  of  the  paper;  take 
C  the  centre  of  this  circle, 
then  C  is  also  the  centre 
of  the  sphere;  draw  the 
diameter  ACE;  and  sup- 
pose another  plane,  per- 
pendicular to  AE,  to  pass  through  C,  the  centre  of  the 
sphere ;  this  plane  will  be  perpendicular  to  the  plane  ABED 
(17.2  sup.),  and  its  common  section  with  the  spherical  surface 
will  be  a  great  circle ;  which  circle,  seen  edgewise,  may  be 
represented  by  BD,  a  diameter  to  the  circle  ABED  at  right 
angles  to  AE.  If,  then,  A  be  taken  as  the  projecting  point, 
the  circle  represented  by  BD  will  be  the  primitive  circle, 
whose  centre  is  C ;  and  as  the  pole  E,  opposite  to  the  pro- 
jecting point,  is  projected  in  C,  the  pole  of  the  primitive 
circle  is  projected  in  its  centre. 


Art.  142.  Any  point  on  the  sphere  is  projected  into  a  point 
distant  from  the  centre  of  the  primitive,  the  semitangent  of 
the  arc  of  a  great  circle  intercepted  between  the  given  point 
and  the  pole  of  the  primitive,  opposite  the  projecting  point. 

Retaining  the  construction  of  the  last  article,  let  F  and  G 
be  two  points  to  be  projected ;  join  AF,  AG ;  and  let  AF, 
AG,  produced  if  necessary,  cut  the  plane  of  the  primitive  in 
I  and  H ;  these  points  are  evidently  the  projections  of  F  and 
G.     But  IC  is  the  tangent  of  CAI,  and  CH  the  tangent  of 


SPHERICAL   PROJECTIONS.  173 

CAH;  that  is,  CI  is  the  semitangent  of  ECF  (20.3)  or  of  EF  ; 
and  CH  the  semitangent  of  ECG,  or  of  EG.  Q.  E.  D. 

Cor.  Any  point  in  the  hemisphere  BED,  opposite  to  the 
projecting  point,  will  be  projected  within  the  primitive  cir- 
cle; but  a  point  on  the  hemisphere  BAD,  adjacent  to  the 
projecting  point,  will  be  projected  without  the  primitive. 
For  the  distance  of  any  point  in  the  first  hemisphere  from 
the  pole  E,  is  less  than  a  quadrant ;  but  in  the  second  it  is 
greater ;  and  the  semitangent  of  90°  =  tangent  of  45°  = 
radius. 

Art.  143.  Every  circle  of  the  sphere  which  passes  through 
the  projecting  point,  is  projected  into  a  right  line  at  right 
angles  to  its  line  of  measures.* 

The  original  circle,  being  in  the  same  plane  as  the  project- 
ing point,  cannot  be  projected  out  of  that  plane;  it  will, 
therefore,  be  projected  into  the  common  section  of  that  plane 
and  the  plane  of  projection :  but  that  section  is  a  right  line 
(3.2  sup.) ;  therefore,  the  projection  of  the  circle  is  a  right 
line.  And  as  the  plane  of  the  circle  projected,  and  the  plane 
of  projection,  are  at  right  angles  to  the  plane  which  forms 
the  line  of  measures  (Art.  136,  Def.  6),  their  common  section 
is  at  right  angles  to  that  plane  (18.2  sup.);  and,  therefore, 
the  line  of  measures  is  at  right  angles  to  the  projection  of  the 
given  circle. 

Cor.  1.  A  circle  which  passes  through  the  poles  of  the 
primitive,  is  projected  into  a  right  line  which  passes  through 
the  centre  of  the  primitive,  at  right  angles  to  its  line  of 
measures. 

Cor.  2.  Every  circle  which  passes  through  the  poles  of  the 
primitive  is  projected  into  a  line  of  semitangents.  This  is 
evident  from  Art.  142. 

*  The  projection  of  a  circle  which  passes  through  the  poles  of  the  primi- 
tive, is  usually  called  a  right  ciiaie.    £v*> 

23 


174 


SECTION    V. 


Art.  144.  Every  circle  of  the  sphere  which  does  not  pass 
through  the  projecting  point,  is  projected  into  a.  circle. 

Case  1.  When  the  plane  of  the  original  circle  is  parallel  to 
the  plane  of  projection. 

Let  A  be  the  projecting  point ; 
E,  the  opposite  pole;  BCD,  the 
plane  of  projection  at  right  angles 
to  AE ;  C,  the  centre  of  the  pri- 
mitive circle ;  FHGI,  the  circle  to 
be  projected;  LMNP,  its  projec- 
tion. If,  now,  while  the  point  A 
remains  fixed,  we  suppose  the  line 
AF  carried  round  with  a  conical 
motion,  so  as  to  describe  the  circle  FHGI ;  the  common  sec^ 
tion  of  the  conical  surface  and  the  plane  of  projection  will 
be  LMNP,  the  projection  of  the  circle  FHGI.  But  the  plane 
of  that  section,  being  parallel  to  the  plane  of  the  base,  is  a 
circle  (Art.  80). 

Cor.  The  radius,  CN  or  CL,  of  the  projected  circle  is  the 
semitangent  of  EG  or  EF,  the  distance  of  the  original  circle 
from  the  pole  opposite  to  the  projecting  point. 

Case  2.  When  the  circle  to  be  projected  is  not  parallel  to 
the  plane  of  projection. 

Let  A  be  the  projecting 
point;  E,  the  opposite  pole; 
FHGIj  the  circle  to  be  project- 
ed ;  LMNP,  its  projection ; 
ABED,  the  common  section 
of  the  spherical  surface  and  a 
plane  which  passes  through 
the  axes  both  of  the  primitive 
and  of  the  circle  FHGI,  and 
therefore  at  right  angles  to  both  these  planes.     Then  BN,  the 


SPHERICAL   PROJECTIONS.  175 

common  section  of  this  plane  and  the  plane  of  projection,  is 
the  line  of  measures  for  the  circle  FHGI  (Art.  136,  Def.  6). 

Supposing,  as  before,  the  line  AF  to  be  carried  round  the 
circle  FHGI,  it  will  describe  a  conical  surface,  whose  com- 
mon section  with  the  plane  of  projection  will  be  LMNP,  the 
projection  of  FHGI. 

Because  the  plane  ABED  passes  through  the  axis  of 
FHGI,  it  must  pass  through  its  centre  and  the  axis  of  the 
cone;  therefore  the  line  FG,  the  common  section  of  this  plane 
and  the  plane  of  the  circle,  is  a  diameter,  which  is  projected 
into  the  line  LN. 

Draw  GK  parallel  to  NB.  Then  the  angle  LNA  =  KGA 
(29.1)  =  AFG  (21.3),  because  AK=AG;  hence  the  triangles 
AFG,  ANL,  which  have  the  angle  at  A  common,  are  equi- 
angular to  each  other ;  and  the  section  LMNP  is  a  subcon- 
trary  section,  and  therefore  (Art.  82)  is  a  circle,  whose 
diameter  is  LN. 

Cor.  The  projected  pole  and  centre  of  the  projected  circle 
are  both  in  the  line  of  measures. 

Art.  145.  The  centre  of  a  projected  less  circle,  at  right 
angles  to  the  primitive,  is  in  the  line  of  measures,  distant 
from  the  centre  of  the  primitive  the  secant  of  the  circle's 
distance  from  its  own  pole ;  and  the  radius  of  the  projected 
circle  is  the  tangent  of  the  same  distance. 

Let  A  (see  fig.  on  p.  176)  be  the  projecting  point ;  ABED,  as 
before,  the  common  section  of  the  spherical  surface  and  a  plane 
which  passes  through  the  centre  of  the  sphere,  and  is  at  right 
angles  both  to  the  plane  of  projection  and  the  plane  of  the  cir- 
cle to  be  projected ;  BCDN,  the  plane  of  projection,  seen  edge- 
wise ;  C,  the  centre  of  the  sphere ;  FG,  the  common  section  of 
the  circle  to  be  projected  and  the  plane  ABED  ;  FG  will  then 
represent  that  circle,  seen  edgewise.  Join  AF,  AG,  CG  and 
EG ;  and  let  AF,  AG  meet  BN  in  L  and  N ;  these  points 
will  then  be  the  projections  of  F  and  G ;  and,  consequently, 


176 


SECTION   V. 
E 


the  line  LN  will  be  the  projection  of  FG,  the  diameter  of  the 
circle  to  be  projected.  As  the  circle  to  be  projected  is  per- 
pendicular to  the  plane  of  projection,  that  plane  must  pass 
through  its  poles ;  hence  the  point  D,  where  that  plane  cuts 
the  circle  ABED,  is  one  of  the  poles ;  and,  therefore,  FD  = 
DG ;  also,  BN  is  the  line  of  measures.  Draw  GP  touching 
the  circle  ABED  in  G,  and  cutting  BN  in  P;  then,  as  proved 
in  Art.  144,  the  angle  ANL  =  AFG.  And  since  GP  touches 
the  circle  AGDE,  and  GA  cuts  it,  the  angle  PGN  =  GFA 
(32.3)  =  GNP;  consequently,  PN  -  PG  (6.1). 

Again,  since  ACD  =  ECD,  being  both  right  angles ;  and 
CAF=CEG;  it  is  plain  (26.1)  that  EG  cuts  CD  in  L;  then, 
since  CG=CE,  the  angle  CGL=CI^L.  Taking  these  equals 
from  the  right  angles  CGP,  LCA;  the  angle  LGP=CLE= 
PLG;  hence,  LP=PG.  Consequently,  PG  —  the  radius  of 
the  circle  described  on  the  diameter  LN;  but  GP  is  the 
tangent,  and  CP  the  secant  of  GD. 

If  now  we  suppose  the  figure  to  revolve  on  BN  until  the 
plane  of  ABED  becomes  perpendicular  to  CA,  the  circle 
ABED  will  be  the  primitive  circle ;  and  the  points  L,  D,  P, 
N,  will  remain  unchanged  :  consequently,  the  circle  described 
from  the  centre  P  with  the  radius  PL  =  PG,  will  be  the  pro- 
jected circle  proposed. 


SPHERICAL   PROJECTIONS. 


it- 


Art.  146.  Any  oblique*  great  circle  will  be  projected  into 
a  circle  whose  centre  is  in  the  line  of  measures,  distant  from 
the  centre  of  the  primitive,  the  tangent  of  its  inclination  to 
the  primitive,  and  the  radius  of  the  projected  circle  is  the 
secant  of  that  inclination. 


Let,  as  before,  A  be  the  projecting  point ;  ABED,  the  great 
circle  at  right  angles  to  the  primitive,  and  to  the  circle  to  be 
projected ;  BN  and  FG,  the  common  sections  of  the  plane  of 
this  circle  writh  the  plane  of  projection,  and  with  the  plane 
of  the  circle  to  be  projected,  respectively.  Then  BN  will 
represent  the  plane  of  projection,  and  FG  the  circle  to  be 
projected,  both  seen  edgewise ;  the  line  BN  will  also  be  the 
line  of  measures.  Join  AF,  AG,  meeting  BN  in  L  and  N ; 
these  will  be  the  projections  of  F  and  G,  and  LN  the  dia- 
meter of  the  projected  circle.  Now,  the  plane  of  the  primi- 
tive circle,  and  of  the  circle  to  be  projected,  being  both  at 
right  angles  to  the  plane  of  ABED,  their  common  section, 
which  passes  through  C,  the  centre  of  the  sphere  (Art.  45),  is  at 
right  angles  to  that  plane  (18.2  sup.) ;  hence  CB  and  CF  are  at 
right  angles  to  that  common  section ;  consequently,  the  angle 
FCB  is  the  inclination  of  the  circle  FG  to  the  primitive  (def. 


*  A  circle  whose  plane  makes  an  oblique  angle  with  the  plane  of  prnjpc 
tion,  is  called  an  oblique  circle. 

23 


173 


SECTION   V. 


4.2  sup).  Draw  AI,  making  the  angle  CAI  ==  FCB.  To 
these  equals  add  CAL  =  CFL,  and  LAI=FLB  (32.1)  =  ILA 
(15.1).     Consequently,  LI  -  AI  (6.1). 

Again,  the  angle  FAG  in  a  semicircle  being  a  right  angle 
(31.3),  is  equal  to  BCF+CAF+CFA  (32.1)  =  LAI  +  CFA. 
Hence,  IAN  =  CFA.  But  ANI^AFC,  by  subcontrary  sec- 
tion (Art.  144) ;  wherefore,  ANI= IAN,  and  AI=IN.  Hence 
the  radius  of  the  circle  described  on  LN,  that  is,  the  projec- 
tion of  FG,  is  equal  to  AI.  But  AI  is  the  secant,  and  CI  the 
tangent,  of  CAI  or  BCF,  the  inclination  of  the  circle  FG  to 
the  primitive. 

If,  then,  as  before,  we  suppose  the  figure  to  revolve  on  BN 
until  the  plane  of  ABED  becomes  perpendicular  to  AC,  the 
circle  ABED  will  be  the  primitive ;  and  the  circle  described 
from  the  centre  I,  with  the  radius  IA  or  IL,  will  be  the  pro- 
jection proposed. 

Cor.  1.  Hence  an  oblique  great  circle  being  projected  on 
the  plane  of  the  primitive;  and,  from  the  point  where  the 
projected  circle  cuts  the  primitive,  two  right  lines  being 
drawn  to  the  centre  of  the  primitive  and  of  the  projected 
circle ;  the  inclination  of  those  lines  is  the  same  as  the  incli- 
nation of  the  original  circle  to  the  primitive. 


SPHERICAL   PROJECTIONS. 


179 


Cor.  2.  Of  all  projected  great  circles,  the  primitive  is  the 
least ;  for  the  secant  of  any  arc  is  greater  than  the  radius. 

Def.  8.  The  angle  made  by  two  circles,  whether  on  the 
same  or  different  planes,  is  the  angle  made  by  their  tangents 
passing  through  the  point  of  intersection. 

When  the  circles  are  both  great  circles,  the  tangents  are 
at  right  angles  to  the  diameter  of  the  sphere,  passing  through 
the  point  of  intersection,  which  is  the  common  section  of  the 
planes  of  these  circles ;  and,  consequently,  the  angle  made  by 
the  tangents  measures  the  inclination  of  the  planes. 

The  definition  contained  in  Art.  45  is  therefore  but  a  par- 
ticular application  of  the  more  general  one  now  given. 

Cor.  The  angle  made  by  the  radii  (drawn  to  the  point  of 
intersection)  of  two  circles  on  the  same  plane,  is  equal  to  the 
angle  made  by  the  circles. 

Art.  147.  The  angle  made  by  two  great  circles  on  the 
surface  of  the  sphere,  is  equal  to  the  angle  made  by  their 
representatives  on  the  plane  of  projection. 

Let  A  be  the  place 
of  the  eye  or  projecting 
point;  B,  the  opposite 
pole;  O,  the  centre  of 
the  sphere ;  HCLG,  the 
primitive  circle ;  I  and 
K,  the  poles  of  the  pro- 
posed great  circles ; 
ACIB,  a  great  circle 
passing  through  A,  B 
and  I ;  BKLA,  another 
great  circle  passing 
through  A,  B  and  K. 

Draw  AOB,  the  axis  of  the  primitive  ;  and  01,  OK,  the  axes 
of  the  proposed  circles.  Let  the  plane  ACIB  cut  the  plane 
of  projection  in  the  line  OC ;  and  BKLA  cut  it  in  OL ;  then 


180  SECTION    V. 

(def.  6),  OC  and  OL  will  be  the  lines  of  measures  of  the  cir- 
cles, whose  poles  are  I  and  K  respectively.  Also  (Cor.,  Art. 
47),  the  angles  IOB,  KOB  and  IOK  are  respectively  equal  to 
the  angles  which  the  proposed  circles  make,  on  the  surface 
of  the  sphere,  with  the  primitive  and  with  each  other.  In 
the  plane  ACIB,  suppose  the  line  AM  to  be  drawn  parallel 
to  OJ,  meeting  OC  in  M.  Then,  since  the  angle  OAM=BOI 
(29.1),  the  angle  which  the  great  circle,  whose  pole  is  I, 
makes  with  the  primitive ;  and  M  is  in  the  line  of  measures 
of  that  circle ;  it  follows  (Art.  146)  that  M  is  the  centre,  and 
MA  the  radius,  of  the  representative  of  that  circle  on  the 
plane  of  projection.  In  like  manner,  suppose  AN,  in  the 
plane  BKLA,  drawn  parallel  to  the  axis  OK,  and  meeting  the 
line  OL  in  N ;  then  N  will  be  the  centre,  and  NA  the  radius, 
of  the  circle  on  the  plane  of  projection  which  represents  the 
original  circle,  whose  pole  is  K.  Now,  since  AM  is  parallel 
to  01,  and  AN  to  OK,  it  follows  (9.2  sup.),  that  the  angle 
MAN  ==  IOK,  the  inclination  of  the  proposed  circles. 

Lastly,  the  points  M,  N,  being  in  the  plane  of  projection, 
let  the  triangle  MAN  revolve  on  MN,  till  the  point  A  falls 
into  the  same  plane,  and  its  position  will  evidently  be  the 
point  where  the  projected  circles  intersect  each  other ;  and 
as  the  radii  of  those  circles  make,  at  the  point  of  intersection, 
the  same  angle  as  the  original  circles  on  the  surface  of  the 
sphere,  the  projected  circles  themselves  make  the  same  angle 
(cor.,  def.  8).  Q.  E.  D. 

Art.  148.  If  a  tangent  to  an  original  circle  be  projected, 
the  projected  tangent  will  be  a  tangent  to  the  projected  cir- 
cle, provided  the  original  circle  does  not  pass  through  the 
projecting  point. 

A  projected  circle  is  the  intersection  of  the  plane  of  the 
primitive  and  a  cone,  whose  vertex  is  the  projecting  point, 
and  base  the  original  circle.  If  a  plane  be  supposed  to  pass 
both  through  the  projecting  point  and  a  tangent  to  the  origi- 
nal circle,  this  plane  will  evidently  touch  the  surface  of  the 


SPHERICAL  PROJECTIONS. 


181 


cone :  and  the  intersection  of  this  plane  and  the  plane  of  pro- 
jection will  be  a  tangent  to  the  intersection  of  the  cone  and 
the  plane  of  projection.  But  the  former  of  these  intersections 
is  the  projected  tangent,  and  the  latter  the  projected  circle. 

Q.  E.  D. 

Cor.  If  two  original  circles  have  a  common  tangent,  the 
projections  of  these  circles  will  have  their  radii  drawn  to  the 
point  of  contact,  in  the  same  straight  line. 

For  the  radii  of  both  the  projected  circles  is  in  a  line 
drawn  through  the  point  of  contact  at  right  angles  to  the 
projected  tangent. 

Art.  149.  The  angle  made  by  any  two  circles  on  the  sur- 
face of  the  sphere,  is  equal  to  the  angle  made  by  their  repre- 
sentatives on  the  plane  of  projection. 

Let  DIBC  be  a  great 
circle  of  the  sphere ;  BT, 
a  straight  line  touching 
it  at  the  point  B ;  through 
BT  let  another  plane 
pass,  cutting  the  sphere 
in  the  circle  BGF;  this  cir- 
cle is,  by  Def.  1,  Art.  45,  a 
less  circle.  As  BT  is  a 
tangent  to  the  great  cir- 
cle, it  is  a  tangent  to  the 
sphere,  and  therefore  to 
the  circle  BGF.  Conse- 
quently, if  these  circles  are  projected,  their  projections  will 
have  their  radii,  which  are  drawn  to  the  point  of  contact,  in 
the  same  straight  line  (Cor.,  Art.  148).  If,  then,  through  the 
point  B,  another  great  circle  and  a  less  one,  having  a  common 
tangent,  be  supposed  to  pass,  these  circles,  when  delineated 
on  the  plane  of  projection,  will  have  their  radii,  which  are 
drawn  to  the  point  of  contact,  also  in  the  same  straight  line. 

24 


182  SECTION   V. 

But  (Art.  147)  the  angle  made  by  two  great  circles  on  the 
surface  of  the  sphere,  is  equal  to  the  angle  made  by  their 
representatives  on  the  plane  of  projection,  or  by  the  radii  of 
those  representatives  drawn  to  the  point  of  intersection.  And 
(def.  8)  the  angle  made  by  two  circles  is  the  angle  made  by 
their  tangents  passing  through  the  point  of  intersection; 
hence  it  is  obvious  that  the  angle  made  by  the  two  less  cir- 
cles, or  by  either  of  them,  with  a  great  circle  touching  the 
other  at  the  point  of  intersection,  is  equal  to  the  angle  made 
by  the  two  great  circles. 

Therefore  the  truth  of  the  proposition  is  manifest. 

Q.  E.  D. 

Art.  150.  The  extremities  of  the  diameter  of  a  projected 
circle  are  in  the  line  of  measures,  distant  from  the  centre  of 
the  primitive,  the  semitangents  of  the  least  and  greatest  dis- 
tances of  the  original  circle  from  the  pole  of  the  primitive 
opposite  to  the  projecting  point.     . 


Let  A  be  the  projecting 
point;  ABED,  the  great  cir- 
cle whose  plane  is  at  right 
angles  to  the  plane  of  the 
primitive  and  of  the  circle  to 
be  projected ;  BD  the  primi- 
tive, and  FG  the  circle  to  be 
projected,  both  seen  edgewise. 
Then  the  right  line  BD,  which 
is  the  common  section  of  the 
plane  ABED  and  the  plane  of  the  primitive,  is  the  line  of 
measures  of  the  circle  FG  (Art.  136,  Def.  6).  As  the  extre- 
mities F  and  G  are  projected  into  H  and  I,  the  line  IH,  which 
is  in  the  line  of  measures  BD,  is  evidently  the  diameter  of 
the  projected  circle.  Also,  E  being  the  pole  opposite  to  the 
projecting  point,  EF  and  EG  are  the  least  and  greatest  dis- 


SPHERICAL   PROJECTIONS. 


183 


tances  of  the  circle  FG  from  that  pole;  and  CH,  CI  are  the 
semitangents  of  EF,  EG  (Art.  142). 

Cor.  1.  The  points  where  a  projected  oblique  great  circle 
cuts  the  line  of  measures,  within  and  without  the  primitive, 
are  distant  from  the  centre  of  the  primitive  the  tangent  and 
cotangent  of  half  the  complement  of  the  inclination  of  the 
original  circle  to  the  plane  of  the  primitive. 

The  angle  BCF  is  the  inclination  of  the  original  circle  FG 
to  the  plane  of  the  primitive  (see  Art.  146) ;  and  therefore 
FCE  is  the  complement  of  that  inclination.  But  CL  is  the 
tangent  of  LAC  =  tangent  of  half  FCE  (20.3).  Also,  since 
FAG  is  a  right  angle  (31.3),  CN,  the  tangent  of  CAG,  is  the 
cotangent  of  half  FCE. 


Cor.  2.  The  centre  of  a  projected  circle  is  in  the  line  of 
measures,  distant  from  the  centre  of  the  primitive,  half  the 
difference  of  the  semitangents  of  the  greatest  and  least  dis- 
tance from  the  pole  opposite  to  the  projecting  point,  when 
the  circle  encompasses  that  pole;  but  half  the  sum  of  the 
semitangents,  when  the  circle  lies  wholly  on  one  side  of  the 
pole.  • 

Art.  151.  Any  circle  and  its  poles  being  projected  on  the 
plane  of  the  primitive,  the  segments  of  the  diameter  inter- 


184 


SECTION    V. 


cepted  between  its  extremities  and  one  projected  pole,  have 
to  each  other  the  same  ratio  as  the  segments  between  the 
same  extremities  and  the  other  pole. 
F    E 


Let  A  be  the  projecting  point ;  ABED,  as  before,  the  great 
circle  at  right  angles  to  the  plane  of  projection  and  of  the 
circle  to  be  projected ;  BD,  the  line  of  measures ;  FG,  the 
common  section  of  the  plane  ABED  and  the  plane  of  the 
circle  to  be  projected ;  P,  Q,  the  poles  of  the  same  circle. 
Then  PQ  is  a  diameter  to  ABED  at  right  angles  to  FG  (Art. 
45,  Cor.  to  Def.  3) ;  and  the  arc  PF  is  equal  to  PG.  Hence 
F,  P,  G,  being  projected  to  H,  p,  I,  the  angle  HAp  =  IAp 
(21.1) ;  consequently, 

\p  :  H/>  : :  IA  :  HA  (3.6). 

Again,  producing  GA  to  K,  and  joining  QF,  the  angle  QAK 
*9  QFG  (22.3  and  13.1)  =  QAF  (21.3),  because  QG  =  QF. 
Hence  (A.  6), 

lq  :  qU  ::  IA  :  AH; 

consequently,  Ip  :  Up  : :  lq  :  ^H. 

Cor.  Hence,  of  the  two  segments  into  which  the  diameter 
is  divided  by  the  projected  pole,  the  greater  is  that  which 
is   more   remote   from    the   centre  of  the  primitive  circle ; 


SPHERICAL  PROJECTIONS. 


185 


and  therefore  the  centre  of  the  projected  circle  is  furthei 
from  the  centre  of  the  primitive,  than  the  projected  pole. 

Art.  152.  The  projected  poles  of  any  circle  are  in  the  line 
of  measures,  within  and  without  the  primitive ;  and  distant 
from  its  centre,  the  tangent  and  cotangent  of  half  its  inclina- 
tion to  the  primitive. 

Retaining  the  construction  of  the  last  article,  the  angle 
ECP  =  the  inclination  of  the  primitive  to  the  circle,  whose 
poles  are  P  and  Q  (Art.  47,  Cor.) ;  and  these  poles  being  in 
the  circle  ABED,  are  projected  to  p  and  q  in  the  line  of  mea- 
sures BD.     But 

Cp  =  tan  CA^  =  tan  ^ECP  (20.3) ; 
and  pAq  being  a  right  angle  (31.3), 

Cq  ==  cotan  Ckp  =  cotan  JECP. 

Cor.  The  projected  pole  of  the  primitive  is  its  centre ;  and 
the  projected  pole  of  a  right  circle  lies  in  the  primitive. 

Art.  153.  If  two  planes  cut  the  sphere,  and  also  intersect 
each  other,  and  from  the  points  where  their  common  section 
meets  the  spherical  surface,  taken  as  poles,  two  circles  be 
described  at  equal  distances  from  those  poles ;  the  arcs  of 
these  circles,  intercepted  between  the  cutting  planes,  on  the 
same  side  of  the  common  section,  are  equal  to  each  other 

Case  1.  When  the  cutting  planes  both  pass  through  the 
centre  of  the  sphere. 

Let  ABDPLG,  ACEP,  be  the 
cutting  planes;  AP,  their  com- 
mon section ;  BC,  DE,  the  inter- 
•  cepted  arcs ;  BF,  CF,  DH,  EH, 
the  common  sections  of  their 
planes  and  the  cutting  planes. 
It  is  to  be  proved  that  BC=DE. 
Because  the  cutting  planes  pass 
24  q* 


186 


SECTION    V. 


through  the  centre  of  the  sphere,  their  common  section  AP  is 
a  diameter  to  each  of  the  circles  formed  by  the  spherical 
surface  and  the  cutting  planes.  And  since  A,  P,  are  the 
poles  of  the  circles  BC  and  DE,  the  line  AP  is  perpendicular 
to  the  planes  of  those  circles  (Art.  45,  Def.  2) ;  therefore, 
AFB,  AFC,  PHD,  PHE  are  right  angles  (def.  1.2  sup.) ; 
whence  FB  is  parallel  to  HD,  and  FC  to  HE  (28.1) ;  conse- 
quently, the  angle  BFC  =  DHE  (9.2  sup).  The  lines  BF, 
CF,  DH,  EH,  are  also  equal,  because  they  are  sines  of  equal 
arcs ;  wherefore,  BC  =  DE  (26.3). 


Case  2.  When  one  of  the  cutting  planes  passes  through 
the  centre  of  the  sphere,  and  the  other  does  not. 

Let  ABDP  be  the  common 
section  of  the  spherical  surface 
and  the  cutting  plane,  which 
passes  through  the  centre; 
ACEP,  the  common  section  of 
the  same  surface  and  the  other 
plane ;  AP,  the  common  sec- 
tion of  the  planes;  BC  and  DE, 
as  before,  the  intercepted  arcs 
of  the  circles  described  from  A  and  P ;  O,  the  centre  of  the 
sphere,  and  consequently  of  ABDP ;  BF,  DH,  the  common 
sections  of  the  planes  of  the  circles  BC,  DE,  and  the  plane 
of  ABDP ;  N,  I,  the  intersections  of  BF,  DH,  with  the  line 
AP.  Join  AO,  PO,  FC,  NC,  HE,  IE,  AC,  PE;  then,  as  in 
the  first  case,  BF,  CF,  DH,  EH,  being  sines  of  equal  arcs, 
are  equal  to  each  other ;  and  AFB,  PHD,  right  angles :  AF, 
PH,  are  also  equal,  being  versed  sines  of  equal  arcs.  Now, 
in  the  triangle  AOP,  the  side  AO  =  OP ;  wherefore,  OAP  = 
OPA;  then,  in  the  triangles  AFN,  PHI,  we  have  AF=PH; 
the  angle  FAN=HPI,  and  AFN- PHI ;  whence  (26.1),  AN 
=  PI,  and  FN=HI.  Then,  in  the  triangles  ANC,  PIE,  we 
have  AC  =  PE  (29.3),  AN  =  PI,  and  the  angle  NAC  =  IPE 


SPHERICAL    PROJECTIONS.  IB* 

(21.3) ;  consequently,  NC  =  IE  (4.1).  Lastly,  in  the  trian- 
gles CFN,  EHI,  the  sides  are  respectively  equal ;  wherefore 
the  angle  NFC  =  IHE  (8.1),  and  consequently  the  arc  BC  — 
DE  (26.3). 

Case  3.  When  neither  of  the  cutting  planes  passes  through 
the  centre  of  the  sphere. 

Through  the  common  section  of  these  planes  and  the  cen- 
tre of  the  sphere,  let  a  third  plane  pass ;  then,  by  the  last 
case,  the  arcs  of  one  of  those  equidistant  circles,  intercepted 
between  the  third  plane  and  each  of  the  others,  are  respect- 
ively equal  to  the  corresponding  arcs  of  the  other  similarly 
intercepted :  and,  therefore,  their  sums  or  differences  are  also 
equal.  But,  when  the  third  plane  passes  between  the  other 
two,  the  sum  of  the  arcs  contained  between  it  and  the  other 
planes  is  the  arc  in  question.  When  it  passes  on  the  same 
side  of  them,  the  difference  is  the  arc  proposed. 

Art.  154.  Let  EFGH,  efgh,  be  the  projections  of  two 
equal  circles,  of  which  EFGH  is  as  far  from  its  pole  P  as 
efgh  is  from  the  projecting  point ;  then  any  two  right  lines 
EP,  FP,  drawn  through  P,  will  intercept  the  representatives 
of  equal  arcs  of  those  circles;  on  the  same  side,  if  P  falls  within 
the  circles ;  but  on  the  contrary  side,  if  it  falls  without ;  that 
is,  EF  =  e/,  and  GH  =£•/?. 


188 


SECTION  V. 


For  (Art.  153)  two  planes  passing  through  the  projecting 
point  and  the  pole  of  the  original  circle,  which  is  represented 
by  EFGH,  will  cut  off  equal  arcs  from  those  circles.  And 
those  planes  will  (Art.  143)  be  projected  into  right  lines, 
which  will  evidently  pass  through  the  projected  pole  P. 

Cor.  1.  Let  a  circle  be  project- 
ed into  a  right  line  EF  at  right 
angles  to  the  line  of  measures 
I  EG;  and  from  C,  the  centre  of 
the  primitive,  let  a  circle  be  de- 
scribed through  P,  the  projected 
pole  of  EF ;  then  any  two  lines 
PE,  PF,  will  cut  off  from  the  circle  an  arc,  ef,  containing 
the  same  number  of  degrees  as  the  arc  which  is  represented 
by  EF.  And  the  arc,  intercepted  between  PE  and  PF,  of 
any  other  circle  which  passes  through  P,  will  contain  the 
same  number  of  degrees. 

For  any  circle  which  is  projected  into  a  right  line,  must 
pass  through  the  projecting  point  (Arts.  143,  144) ;  and, 
therefore,  the  distance  of  that  circle  from  its  pole  is  the  same 
as  the  distance  of  the  pole  from  the  projecting  point.  Con- 
sequently, the  projected  circle  through  P  represents  an  origi- 
nal circle,  as  far  from  the  projecting  point  as  the  circle  which 
is  projected  into  EF  is  from  its  own  pole.  Hence,  EF  and 
ef  represent  equal  arcs.  The  latter  part  of  the  corollary  is 
evident  from  26.3. 

Cor.  2.  If  two  right  lines  be  drawn  through  the  projected 


SPHERICAL    PROJECTIONS.  189 

pole  of  a  great  circle,  the  intercepted  arc  of  that  circle  will 
contain  the  same  number  of  degrees  as  the  intercepted  arc 
of  the  primitive. 

For  any  great  circle  is  distant  90°  from  its  pole ;  and  the 
primitive  is  90°  from  the  projecting  point. 

Cor.  3.  If,  from  the  point  where  two  projected  great  cir- 
cles cut  each  other,  two  right  lines  are  drawn  through  the 
projected  poles  of  those  circles,  the  intercepted  arc  of  the 
primitive  circle  will  measure  the  spherical  angle  made  by 
those  circles  at  the  point  of  their  inter  sec  ti<*i. 

For  the  arc  of  a  great  circle,  contained  between  the  poles 
of  two  other  great  circles,  is  the  measure  of  the  angle  which 
the  axes  of  these  circles  make  with  each  other ;  and  that 
angle  is  the  same  as  the  inclination  of  the  planes  of  those 
circles  (Cor.,  Art.  47). 

Scholium.  If  the  circles  of  the  sphere  were  to  be  projected 
on  a  plane  parallel  to  the  plane  of  the  primitive,  the  projec- 
tions would  be  similar  to  those  on  the  plane  of  the  primitive 
itself;  for  the  projecting  line,  when  carried  round  on  the  cir- 
cumference of  a  circle  which  does  not  pass  through  the 
projecting  point,  forms  a  conical  surface ;  and  that  surface 
being  cut  by  the  plane  of  the  primitive,  and  by  any  other 
p-lane  parallel  thereto,  the  sections  are  similar,  but  of  greater 
or  less  dimensions.  Thus  projecting  on  the  plane  of  a  less 
circle  parallel  to  the  primitive,  instead  of  projecting  on  the 
primitive  itself,  would  be  only  changing  the  scale.  In  the 
subsequent  parts,  however,  of  this  section,  the  plane  of  the 
primitive  will  be  used. 

In  the  following  problems,  the  primitive  circle  is  always 
supposed  to  be  described  with  the  chord  of  60° ;  and  the 
secant,  tangent,  and  semitangent  referred  to,  are  such  as 
correspond  to  the  scale  used  for  the  primitive.  These  differ- 
ent lines  are  frequently  marked  by  the  side  of  the  scale  of 
chords,  on  the  small  scales  introduced  into  boxes  of  mathe- 

25 


190 


SECTION   V. 


matical  instruments ;  but  more  frequently  on  the  foot  or  two 
feet  scales,  which  contain  Gunter's  lines. 

Art.  155.  Problem.  To  describe  a  circle  parallel  to  the 
primitive,  at  a  given  number  of  degrees  from  its  pole. 

E 

From  the  centre  of  the 
primitive,  with  a  radius  equal 
to  the  semitangent  of  the 
given  distance  of  the  circle 
from  its  own  pole,  describe 
the  circle  required.  Or  draw 
the  diameters  AB,  DE  at 
right  angles  to  each  other; 
and  from  the  extremity  E 
of  one  of  them,  lay  off  the 
proposed  number  of  degrees  on  the  primitive  as  EF ;  join 
DF,  cutting  AB  in  G;  from  the  centre  C,  at  the  distance 
CG,  describe  the  circle  required. 

The  radius  CG  is  the  tangent  of  CDG,  or  semitangent  of 
EF,  as  it  ought  to  be  (Cor.,  Case  1,  Art.  144). 


SPHERICAL   PROJECTIONS. 


191 


Art.  156.  Prob.  To  describe  a  less  circle  at  right  angles 
to  the  primitive,  and  at  a  given  distance  from  its  own  pole. 

Let  B  be  the  pole  of  the  circle  proposed ;  through  C,  the 
centre  of  the  primitive,  and  the  pole  B,  draw  the  right  line 
CD ;  from  C  to  D,  lay  down  the  secant  of  the  given  distance ; 
from  the  centre  D,  with  a  radius  equal  to  the  tangent  of  the 
same  distance,  describe  the  circle  proposed.  Or,  from  B,  lay- 
down  BE  on  the  primitive ;  join  CE ;  draw  ED  touching  the 
circle;  from  D,  with  the  distance  DE,  describe  the  circle 
proposed. 

It  is  obvious  that  DE  is  the  tangent,  and  CD  the  secant, 
of  BE,  as  they  ought  to  be,  Art.  145. 

Art.  157.  Trob.  To  describe  an  oblique  circle  at  a  given 
distance  from  a  given  pole. 

hetp  be  the  projected 
pole ;  through  p  draw 
the  line  of  measures 
ApB ;  apply  C/>,  the  dis- 
tance of  the  given  pole 
from  the  centre  of  the 
primitive,  to  the  line 
Qf  semitangents ;  and, 
having  found  the  num- 
ber of  degrees,  thus 
measured,  in  Cp,  take  the  sum  and  difference  of  this  number 
and  the  distance  of  the  proposed  circle  from  its  own  pole ; 
lay  down  these  results  taken  from  the  semitangents,  on  the 
line  of  measures,  from  C  to  g  and/;  on  the  diameter  fg  de- 
scribe the  circle  required.  Or,  having  drawn  the  line  of 
measures,  draw  the  diameter  DCE  at  right  angles  to  it; 
draw  Dp  to  meet  the  primitive  in  P ;  from  P,  lay  down  on 
the  primitive  PF,  PG,  each  equal  to  the  given  distance  of  the 
circle  from  its  pole ;  draw  DF,  DG,  cutting  the  line  of  mea- 
sures in /and  g;  on  fg  describe  a  circle ;  it  will  be  the  circle 
required. 


192  SECTION   V. 

This  construction  follows  from  Art.  150. 

Scholium.  This  method  is  applicable  to  great  circles  as 
well  as  less ;  but  the  former  cases  are  conveniently  managed 
by  other  methods  hereafter  given. 

Art.  158.  Prob.  To  describe  a  great  circle,  the  projected 
pole  of  which  is  given  in  position. 

Case  1.  When  the  given  pole  is  in  the  primitive  circle. 

Through  the  given  pole  draw  the  line  of  measures ;  and  at 
right  angles  thereto  draw  a  diameter  to  the  primitive  circle ; 
this  diameter  will  represent  the  circle  proposed. 

Because  the  pole  is  in  the  primitive,  the  original  circle  is 
at  right  angles  to  the  primitive  (Art.  45,  Cor..  Def.  3) ;  and, 
being  a  great  circle,  it  must  pass  through  the  poles  of  the 
primitive.  Consequently  (Art.  143,  Cor.  1),  it  is  represented 
by  a  right  line  through  the  centre  of  the  primitive,  at  right 
angles  to  the  line  of  measures. 


Case  2.  When 
circle. 


the   given   pole   is  within   the   primitive 


Let  p  be  the  projected  pole ; 
through  p  draw  the  line  of 
measures  CpG;  apply  Cp  to 
the  line  of  semitangents ;  take 
CG  equal  to  the  tangent  of  the 
number  of  degrees  in  Cp; 
from  G,  as  a  centre,  with  the 
seGant  of  the  same  number 
of  degrees,  describe  the  circle 
D  DHE,  which  will  be  the  one 

required. 

Or,  draw  the  diameter  DCE  at  right  angles  to  the  line  of 
measures ;  join  Dp,  and  produce  it  to  the  circumference  in 
P  ;  make  PF  =  PE ;  draw  DF  cutting  the  line  of  measures 
in  G ;  from  the  centre  G,  with  the  radius  GD,  describe  the 


SPHERICAL   PROJECTIONS. 


193 


circle  DHE,  the  circle  proposed.  The  line  Cp  is  the  projec- 
tion of  an  arc  of  a  great  circle,  intercepted  between  the  pole 
of  the  primitive  and  the  pole  of  the  circle  proposed  ;  and  that 
arc  measures  the  inclination  of  those  circles  (Art.  47,  Cor). 
The  arc  is  also  projected  into  a  line  of  semitangents  (Art. 
143,  Cor.  2) ;  hence  Cp,  measured  on  the  semitangents,  or  the 
arc  PE,  indicates  the  same  inclination.  But,  by  the  con- 
struction, CG  is  the  tangent  and  GD  the  secant  of  PE ;  con- 
sequently (Art.  146),  DHE  is  the  circle,  whose  pole  is  j>. 


Art.  159.  Prob.  Through  two  given  points,  to  describe  a 
great  circle. 

Let  A,  B  be  the 
given  points ;  through 
the  centre  of  the  pri- 
mitive and  one  of  the 
given  points  draw  the 
right  line  ACF.  If 
that  line  passes  through 
B,  the  business  is  done ; 
for  ACF  is  the  projec- 
tion of  a  great  circle 
at  right  angles  to  the  primitive  (Art.  143,  Cor.  1).  But  if 
ACF  does  not  pass  through  B,  apply  CA  to  the  line  of  semi- 
tangents, and  make  CF  the  semitangent  of  the  supplement 
of  CA ;  through  ABF  describe  a  circle,  and  the  thing  is  done. 
Or,  draw  the  diameter  DCE  at  right  angles  to  ACF ;  draw 
DAG  cutting  the  primitive  in  G ;  draw  the  diameter  GCH ; 
join  DH ;  and  let  DH,  produced  if  necessary,  cut  AF  in  F ; 
and  through  ABF  describe  the  circle  ABF,  as  before. 

From  the  construction,  it  is  obvious  that  AF  is  the  projec- 
tion of  a  semicircle ;  consequently,  any  circle  which  passes 
through  A  and  F  must  be  a  great  one  (converse  to  Cor.  3, 
Def.  1,  Art.  45). 
25 


194 


SECTION    V. 


Art.  160.  Prob.  About  a  given  pole,  to  describe  a  circle 
through  a  given  point. 

Let  p  be  the  given  pole 
and  B  the  given  point ; 
through  p  and  B  describe 
a  great  circle  (Art.  159) ; 
and  draw  BF  touching  it 
in  B  (17.3) ;  draw,  from 
the  centre  of  the  primi- 
tive, the  right  line  C/)F, 
meeting  the  tangent  in  F ; 
from  the  centre  F,  at  the 
distance  FB,  describe  the  circle  BGH ;  and  the  work  is  done. 
The  centre  of  a  circle  whose  pole  is  p  is  in  the  line  of  mea- 
sures QoF  (Cor.  to  Case  2,  Art.  144).  The  circle  />B,  passing 
through  the  pole  p  of  the  proposed  circle,  is  at  right  angles 
to  it  (Cor.  to  Def.  3,  Art.  45) ;  hence  the  radii  of  their  pro- 
jections, drawn  to  the  point  of  their  intersection,  must  also 
be  at  right  angles  to  each  other  (Art.  149) ;  consequently,  the 
centre  of  the  required  circle  is  in  the  tangent  BF  (18.3) ;  it 
is,  therefore,  at  the  intersection  of  Cp  and  BF. 

Art.  161.  Prob.  To  find  the  poles  of  a  given  projected 
circle  FNG. 

E 


Through  the  centre  of  the  given  circle  and  centre  C  of  the 
primitive,  draw  the  right  line  FCGP,  cutting  the  given  circle 


SPHERICAL   PROJECTIONS. 


135 


in  F  and  G.  Measure  CF  and  CG  on  the  line  of  semitan- 
gents  ;  take  the  half  sum  or  half  difference  of  these  measures, 
according  as  F  and  G  are  on  the  same  or  opposite  sides  of  C, 
and  lay  its  semitangent  from  C  to  p ;  then  is  p  one  of  the 
poles  required :  observing,  however,  that  p  must  be  on  the 
same  side  of  C  as  the  centre  of  the  given  circle.  Lay  down 
CP  from  the  semitangents  equal  to  the  supplement  of  Cp,  and 
on  the  opposite  side  of  C ;  then  is  P  the  other  pole  required. 
Or,  draw  the  diameter  ACE  at  right  angles  to  FG ;  draw 
also  AG,  AF,  cutting  the  primitive  in  L  and  M ;  bisect  LM 
in  I ;  join  AI,  cutting  FG  in  p ;  draw  the  diameter  IH  to  the 
primitive  circle ;  draw  AH,  produced  if  necessary,  to  meet 
FG  produced  in  P ;  then  p  and  P  are  the  poles  required. 

When  the  circle  is  a  great  one,  as  AGE,  the  poles  may  be 
found  with  more  facility  in  a  different  manner. 


Draw  FC  from  the  centre  of  the  given  circle  to  the  centre 
of  the  primitive,  and  produce  it.  Measure  CF  on  the  line 
of  tangents:  take  half  the  number  of  degrees  thus  found,  and 
lay  down  the  tangent  of  this  result  from  C  towards  F  to  p ; 
and  its  cotangent  in  the  opposite  direction  to  P ;  then  will  p 
and  P  be  the  poles  required.  Or,  draw  the  diameter  ACE 
at  right  angles  to  FC  ;  join  AF ;  bisect  the  angle  CAF  by  the 
line  Apl,  cutting  FC  in  p  and  the  primitive  circle  in  I ;  draw 
the  diameter  IH  and  the  line  AHP  as  before ;  then  p  and  P 
are  the  poles  required. 


196 


SECTION    V. 
E 


In  the  case  of  the  less  circle,  CG  is  the  semitangent  of 
EL,  and  CF  the  semitangent  of  EM;  consequently  (i\.rt. 
150)  EL  and  EM  are  equal  to  the  least  and  greatest  dis- 
tances of  the  original  circle  represented  by  FNG,  from  the 
pole  of  the  primitive  opposite  the  projecting  point.  Hence 
EI,  half  the  difference  of  EM  and  EL,  when  F  and  G  are  on 
opposite  sides  of  C  (as  in  the  figure),  or  half  the  sum  when 
F  and  G  are  on  the  same  side,  is  the  distance  of  the  pole  of 
the  original  circle  from  the  pole  of  the  primitive.  Cp,  the 
semitangent  of  EI,  is  therefore  the  distance  of  the  projected 
pole  of  FNG  from  the  centre  of  the  primitive  (Art.  142). 
And  the  projected  pole  lies  in  the  line  which  joins  the  centre 
of  the  projected  circle  and  the  centre  of  the  primitive  (Art. 
144,  Cor.  to  Case  2).  Also,  p  and  P  are  on  the  same  project- 
ed great  circle,  at  the  distance  of  180°;  hence,  P  is  the  pole 
opposite  to  p. 

In  the  case  of  the  great  circle,  CF  is  the  tangent  of  CAF, 
the  inclination  of  AGE  to  the  primitive  (Art.  146,  Cor.  1). 
But,  by  the  construction,  ECI  =  CAF ;  consequently,  the  arc 
EI  measures  the  distance  of  the  pole  of  the  original  circle, 
represented  by  AGE,  from  the  pole  of  the  primitive  (Cor.  to 
Art.  47).  Hence  Cp,  the  semitangent  of  EI,  is  equal  to  the 
distance  of  the  projected  pole  of  AGE  from  the  centre  of  the 
primitive  (Art.  142). 


SPHERICAL  PROJECTIONS. 

E 


197 


Art.  162.  Prob.  To  describe  a  great  circle  making  a  given 
angle  with  the  primitive  at  a  given  point  A. 

Draw  through  the  given  point  the  diameter  ACE ;  and 
through  C,  the  centre  of  the  primitive  circle,  draw  a  line  CF 
at  right  angles  to  AE ;  make  the  angle  CAF  =  the  angle 
proposed ;  and  from  F,  the  intersection  of  AF,  and  CF,  de- 
scribe the  circle  AGE ;  which  will  be  the  circle  required. 
Or,  from  the  centre  of  the  primitive,  with  the  tangent  of  the 
given  angle,  describe  an  arc;  from  the  point  A,  with  the 
secant  of  the  same  angle,  describe  an  arc  cutting  the  former  • 
and  from  the  point  of  intersection  describe,  through  the  point 
A,  the  circle  AGE. 

This  construction  is  obvious  from  Art.  146. 

When  the  given  angle  is  a  right  one,  the  lines  AF  and  CF 
are  parallel ;  hence,  in  that  case,  the  centre  of  the  required 
circle  is  at  an  infinite  distance ;  consequently,  the  circle  be- 
comes a  right  line  passing  through  the  centre  of  the  primitive. 
See  Cor.  1,  Art.  143. 

Art.  163.  Prob.  Through  a  given  point  P,  to  describe  a 
great  circle,  making  a  given  angle  with  the  primitive. 

From  the  centre  of  the  primitive,  with  the  tangent  of  the 
given  angle,  describe  an  arc;  from  the  given  point,  with  the 
secant  of  the  same  angle,  describe  an  arc  cutting  the  former 
in  F ;  from  the  centre  F,  through  P,  describe  the  circle  APB ; 
this  will  be  the  circle  required. 

26 


198 


SECTION    V. 


Or,  through  P  draw  the 
diameter  DE ;  at  the  cen- 
tre C  erect  a  perpendicular 
CG ;  make  the  angle  CDG 
=  the  given  angle;  from 
C,  through  G,  describe  an 
arc ;  and  from  P,  with  the 
distance  DG,  describe  an- 
other, cutting  the  former 
in  F  ;  then  F  is  the  centre 
of  the  required  circle. 

This,  like  the  last,  de- 
pends upon  Art.  146. 

N.  B.  If  the  circles  described  from  C  and  P  do  not  meet, 
the  problem  is  then  impossible ;  and  this  case  occurs  when 
the  required  angle  is  less  than  that  which  would  be  measured 
by  PD,  taken  on  the  scale  of  semitangents  from  90°  towards 
the  beginning  of  the  scale. 


Art.  164.  Prob.  To  describe  a  great  circle,  making  at  a 
given  point  P  a  given  angle  with  a  given  great  circle  APB. 

Through  the  given  point  P  draw  the  diameter  DE,  meeting 
the  circle  again  in  H;  find  F  the  centre  of  the  given  circle; 
draw  FI  at  right  angles  to  DE  ;  join  PF ;  make  the  angle 
FPL  =  the  given  angle ;  then,  from  L,  describe  the  circle 
MPN ;  this  is  the  circle  required. 

The  line  FI,  being  at  right  angles  to  PH,  bisects  it  (3.3) ; 
consequently,  every  circle  whose  centre  is  in  FI,  and  which 
passes  through  P,  will  also  pass  through  H.  Now,  as  APB 
and  DCE  are  the  projections  of  great  circles,  PBH  is  the 
projection  of  a  semicircle  (Cor.  to  Def.  3,  Art.  45) ;  hence, 
any  other  circle  passing  through  P  and  H  must  be  a  great 
circle.  MPN  is  therefore  a  great  circle ;  also  (Art.  147)  the 
angle  BPN  =  FPL. 


SPHERICAL   PROJECTIONS.  199 

Art.  165.  Prob.  Through  a  given  point  P,  to  describe  a  great 
circle  making  a  given  angle  with  a  given  great  circle  DE. 

Find  the  pole  of  the  given 
circle  (Art.  161);  about  that 
pole  describe  a  circle  GFH 
at  a  distance  equal  to  the 
measure  of  the  given  angle 
(Arts.  155-6-7);  about  the 
point  P  as  a  pole,  describe  a 
great  circle  IFK,  cutting  GFH 
in  F  and  K  (Art.  158) ;  about 
D  one  of  those  points    F   as  a 

pole,  describe  the  great  circle  LPM;  and  the  work  is  done. 

Since  P  is  the  pole  of  IFK,  every  point  in  that  circle  is  90° 
from  P ;  consequently,  the  great  circle  whose  pole  F  is  in 
IFK,  must  pass  through  P.  And  as  the  distance  of  the  poles 
of  two  great  circles  is  the  measure  of  the  angle  which  those 
circles  make  with  each  other,  the  construction  is  manifest. 

If  the  given  angle  is  a  right  one,  the  circle  must  be  de- 
scribed through  P  and  the  pole  of  DE,  by  Art.  159. 

Art.  166.  Prob.  To  describe  a  great  circle  making  given 
angles  with  two  given  great  circles  AB  and  CD.  See  fig.  on 
page  200. 

Find  the  poles  r,  s  of  the  given  circles  (Art.  161) ;  describe 
about  r  and  ft  less  circles,  at  distances  respectively  equal  to 
the  measures  of  the  given  angles  (Art.  157) ;  from  the  inter- 
section E  of  these  circles,  as  a  pole,  describe  the  great  circle 
FG  (Art.  158) ;  that  circle  is  the  one  required. 

This  construction  evidently  depends  upon  the  principle, 
that  the  distance  between  the  poles  of  two  great  circles  is 
the  measure  of  their  inclination  (Cor.  to  Art.  47). 

If  the  circle  to  be  described  is  to  be  at  right  angles  to  each 
of  those  which  are  given,  it  must  be  described  through  their 
poles,  by  Art.  159. 


in 


SECTION   V. 


Art.  167.  Prob.  To  describe  a  right  circle  (that  is,  a  great 
circle  at  right  angles  to  the  primitive)  making  a  given  angle 
with  a  given  great  circle  CD. 

Find  s,  the  pole  of  CD ;  about  s  describe  a  circle  at  a  dis- 
tance equal  to  the  measure  of  the  given  angle ;  from  the  point 
H,  where  this  circle  cuts  the  primitive,  lay  down  HI,  on  the 
primitive,  =  90° ;  through  I  draw  the  diameter  IL,  the  right 
circle  required. 

Every  great  circle  at  right  angles  to  the  primitive  is  pro- 
jected into  a  right  line  through  its  centre  (Art.  143,  Cor.  1). 

[Note.  There  is  a  limit  in  this  and  the  last  article.  If,  in 
Art.  166,  the  circles  about  r  and  s  do  not  meet ;  or,  in  this 
article,  if  the  circle  about  s  does  not  meet  the  primitive ;  the 
problem  is  impossible.] 

Scholium.  When  the  proposed  angle  is  a  right  one,  lay  down 
90'  from  C  on  the  primitive  circle ;  and  through  the  point 
thus  found  draw  a  diameter  for  the  right  circle  required. 

Art.  168.  Prob.  Through  a  given  point  Z,  to  describe  a 
great  circle  which  shall  touch  a  given  less  circle  ABC. 
From  Z,  as  a  pole,  describe  the  great  circle  DGE  (Art 


SPHERICAL   PROJECTIONS. 


201 


158) ;  find  P  the  internal  pole  of  ABC  (Art.  161),  and  Q  the 

opposite  pole ;  about  Q  de- 
scribe a  circle  FGH,  at  a  dis- 
tance from  Q  equal  to  the 
complement  of  PB,  the  dis- 
tance of  ABC  from  its  own 
pole  (Arts.  155-6-7);  from 
the  point  G,  where  these  cir- 
cles cut  each  other,  taken  as 
a  pole,  describe  the  great  cir- 
cle ZIL  (Art.  158) ;  this  cir- 
cle will  touch  the  given  circle 
ABC.  Through  PGQ  de- 
scribe a  circle  cutting  ABC  in  I ;  this  will  be  a  great  circle, 
because  P  and  Q  are  opposite  poles,  and  PI  4-  QG  are  by- 
construction  =  90° ;  hence  GI  ===  90° :  and  G  being  the  pole 
of  ZIL,  that  circle  must  pass  through  I,  and  make  the  angle 
ZIP  a  right  angle.  Hence  (Art.  55)  PI  is  less  than  any  other 
arc  of  a  great  circle  contained  between  P  and  ZIL ;  there- 
fore, those  circles  touch  each  other  at  I. 

Art.  169.  Prob.  To  lay  down  a  given  number  of  degrees 
on  a  given  great  circle,  or  to  measure  an  arc  of  it. 

Case  1.  When  the  given  circle  is  the  primitive,  lay  down 
or  measure  the  arc  by  the  scale  of  chords. 

For  the  primitive  is  an  original  circle,  and  is  therefore 
measured  as  in  common  Geometry. 

Case  2.  When  the  given  circle  is  a  right  one,  that  is,  one 
passing  through  the  centre  of  the  primitive,  lay  down  or 
measure  the  arc  on  the  scale  of  semitangents  (Art.  143,  Cor. 
2),  observing  that  an  arc  beginning  at  the  centre  of  the  pri- 
mitive must  be  measured  from  the  beginning  of  the  scale ; 
but  one  beginning  at  the  primitive  must  be  measured  from 
the  90th  degree  on  the  scale  towards  the  beginning  or  end 
of  the  scale,  according  as  the  arc  extends  towards  or  from 
the  centre  of  the  primitive. 
26 


202 


SECTION  V. 


Or,  let  ACB  be  the  right  cir- 
cle; draw  the  diameter  DCE 
at  right  angles  to  AB ;  then,  to 
lay  down  any  proposed  number 
of  degrees  from  A  or  C,  lay 
them  on  the  primitive  from  A 
or  E  to  F  ;  join  DF,  cutting 
AB  in  G ;  or,  to  measure  AG 
or  CG,  draw  DG  to  cut  the 
primitive  in  F ;  then  AF  is  the 
measure  of  AG,  and  EF  of  CG. 

Suppose  the  figure  to  revolve  on  AB  till  CD  becomes  per- 
pendicular to  the  plane  of  projection;  then  is  D  the  project- 
ing point,  and  AEB  the  semicircle  passing  through  the  pole 
of  the  primitive ;  consequently,  AG  is  the  projection  of  AF, 
and  CG  of  EF. 

Case  3.  When  the  given  circle  is  an  oblique  one. 

Let  DUE  be  the  circle ;  find  its  internal  pole  I  (Art.  161) ; 
then,  to  lay  down  an  arc  HL  or  EL,  lay  the  proposed  num- 
ber of  degrees  on  the  primitive  from  A  or  E  to  F,  and  join  IF, 
cutting  the  given  circle  in  L,  the  point  required ;  or,  to  mea- 
sure HL  or  EL,  join  IL,  and  produce  it  to  F  in  the  primitive ; 
then  AF  is  the  measure  of  HL,  and  EF  of  EL. 

The  primitive  circle  is  as  far  from  the  projecting  point,  as 
DHE  is  from  its  pole ;  therefore  (Art.  154),  the  right  lines 
IA,  IF,  cut  off  corresponding  arcs  AF,  HL. 

Art.  170.  Prob.  To  lay  down  any  proposed  number  of 
degrees  on  a  less  circle,  or  to  measure  a  given  arc  of  it. 

Case  1.  When  the  given  less  circle  is  parallel  to  the 
primitive. 

Lay  the  proposed  number  of  degrees  on  the  primitive 
circle;  and  through  the  extremities  of  the  arc  draw  right 
lines  to  the  centre ;  the  intercepted  arc  of  the  less  circle  is 


SPHERICAL   PROJECTIONS. 


203 


that  proposed  ;  or,  to  measure  the  arc,  draw  right  lines  from 

the  centre,  through  its  extremi- 
ties, to  the  primitive,  and  mea- 
sure the  intercepted  arc  of  the 
latter.  Thus  BH  is  the  measure 
of  IL,  and  EH  of  ML ;  for  the 
B  projected  less  circle  parallel  to 
the  primitive  is  formed  by  cut- 
ting the  conical  surface  by  a 
plane  parallel  to  the  base ;  con- 
sequently, the  projected  circle 
differs  from  its  original  in  nothing 
but  its  dimensions. 

Case  2.  When  the  circle  is  not  parallel  to  the  primitive. 

Let  ABC  be  the  less  cir- 
cle; find  its  pole  D  (Art. 
161) ;  describe  a  circle 
FGH,  as  far  from  the  pro- 
jecting point  as  ABC  is 
from  its  pole  (Art.  155) ; 
then  any  arc  of  ABC  may 
be  laid  down  or  measured 
by  the  aid  of  FGH  as  an 
arc  of  a  great  circle  is,  by 
means  of  the  primitive  in  Art.  169,  Case  3 ;  observing  that 
an  arc  of  FGH  is  laid  down  or  measured  as  directed  in  Case 
1.  Thus,  I  being  the  centre  of  the  primitive,  ML,  a  part  of 
its  circumference,  is  the  measure  of  FG ;  and  FG  the  mea- 
sure of  AE  (Art.  154). 


Art.  171.  To  measure  the  angle  made  by  two  great  circles 
whose  position  is  given. 

Find  the  poles  of  the  circles  (Art.  161) ;  from  the  angular 
point,  through  those  poles,  draw  two  right  lines;  and  the 
intercepted  arc  of  the  primitive  is  the  measure  required. 


204 


SECTION  V. 

Let  ACB,  ECF,  be  the  circles ; 
G,  I,  their  poles ;  then  the  arc  of 
a  great  circle,  contained  between 
I  and  G,  would  measure  the  angle 
ACF  (Art.  154,  Cor.  3).  Or,  draw 
lines  from  the  point  of  intersection 
C  to  the  centres  of  the  circles  ACB 
and  ECF;  then  the  angle  con- 
tained between  these  lines  will 
measure  the  angle  ACF  (Cor., 
Def.8). 


Art.  172.  Prob.  To  form  a  general  projection  of  the  sphere 
on  the  plane  of  a  meridian.* 

Let  ZONH  denote  the  me- 
ridian ;  Z,  the  zenith ;  N,  the 
nadir;  P,  S,  the  north  and 
south  poles;  EQ,  the  equator; 
HO,  the  horizon ;  then  ZE  = 
OP,  the  latitude  of  the  place. 
Then,  circles  being  described 
through  P  and  S,  making  suc- 
cessively angles  of  15°,  30°, 
45°,  60°,  75°  and  90°,  with 
the  primitive;  these  will  be 
the  meridians,  or  hour  circles,  for  the  different  hours. 


A  few  examples  will  now  be  given  to  exercise  the  student 
in  Spherical  Projections  and  Calculations. 


Of  Rectangular  Spherical  Triangles. 

1.  Given,  the  hypothenuse  70°  15',  and  the  adjacent  angle 
30°  30',  to  find  the  rest. 

*  See  Definitions,  page  217. 


SPHERICAL   PROJECTIONS. 


205 


Construction.  Describe 
the  primitive  circle  ABC, 
and  the  oblique  great 
circle  ADC,  making  the 
given  angle  with  the  pri- 
mitive at  the  point  A 
(Art.  162);  on  AD  lay 
AE  equal  to  the  given 
hypothenuse  (Art.  169) ; 
and  through  F,  the  pole 
of  the  primitive,  and  the 
point  E,  describe  the 
great  circle  FEG,  cutting 

the  primitive  circle  in  G;  then  AGE  is  the  triangle  proposed, 

of  which  G  is  the  right  angle. 

Calculation. 

As  rad  :  cos  A  : :  tan  AE  :  tan  AG  (Art.  62)  =  67°  23'. 

As  rad  :  sin  A  : :  sin  AE  :  sin  EG  (Art.  58)  F  28°  32'. 

As  rad  :  cos  AE  : :  tan  A  :  cot  AEG  (Art.  64)  =  78°  45'. 

It  is  obvious,  from  the  construction,  that  there  is  no  ambi- 
guity in  this  problem ;  for  the  points  E  and  F  being  given, 
the  great  circle  passing  through  them  can  have  but  one 
position. 

Again,  the  side  EG  is  of  the  same  affection  as  the  angle  A 
(Art.  56) ;  also,  AG  is  of  the  same  affection  with  EG,  or  a 
different  one,  according  as  AE  is  less  or  greater  than  a  quad- 
rant (Art.  57) ;  and  the  angle  AEG  is  of  the  same  affection 
as  AG  (Art?  56). 


2.  Given,  the  hypothenuse  125°  25',  and  one  leg  37°  40',  to 
find  the  rest. 

Construction.  Having  described  the  primitive  ^:icle  ABC, 
lay  AG  on  it   equal  the   given   leg;    through   G  and   the 

27 


206 


SECTION   V. 


pole  F,  draw  the  right 
circle  GFE ;  from  the 
point  A,  as  a  pole,  at  a 
distance  equal  to  the  hy- 
pothenuse  (or  from  the 
opposite  pole  C  with  its 
supplement),  describe  a 
less  circle  DEH  (by  Art. 
156),  cutting  GFE  in  E ; 
through  A  and  E  describe 
the  great  circle  AEC 
(Art.  159) ;  then  AGE  is 
the  triangle  proposed. 

Calculation. 

rad  :  cos  EAG  (Art.  62)  =  123°  18'. 
rad  :  sin  AEG  (Art.  58)-  48°  34'. 
rad  :  cos  EG  (Art.  65)  =  137°  4'. 
In  this  problem  there  is  no  ambiguity ;  for  the  angle  AEG 
is  of  the  same  affection  with  the  side  AG  (Art.  56) ;  EG  is 
of  the  same  affection  with  AG,  when*AE  is  less  than  a  quad- 
rant, and  of  a  different  one  when  AE  is  greater  (Art.  57) ; 
and  the  angle  EAG  is  of  the  same  affection  as  EG  (Art.  56). 


As  tan  AE 
As  sin  AE 
As  cos  AG 


tan  AG 
sin  AG  : 
:  cos  AE 


3.  Given,  one  leg  75° 
26',  and  the  adjacent  an- 
gle 40°  10',  to  find  the 
rest. 

Construction.  Describe 
the  primitive  circle  ABC, 
and  the  oblique  great  cir- 
cle AEC,  making  BAC 
equal  to  the  given  angle 
(by  Art.  162);  on  the 
primitive  lay  AG  equal 
the  given  leg;  and  through 


SPHERICAL   PROJECTIONS. 


207 


G  and  the  pole  of  the  primitive,  draw  the  right  circle  GEF, 
cutting  AC  in  E ;  then  is  AGE  the  triangle  in  question. 

Calculation. 
As  rad  :  sin  AG  : :  tan  A  :  tan  EG  (Art.  60)  =  39°  15'. 
As  cos  A  :  rad  : :  tan  AG  :  tan  AE  (Art.  62)  =  78°  46'. 
As  rad  :  sin  A  : :  cos  AG  :  cos  AEG  (Art.  67)  =  80°  40'. 

This  problem  includes  no  ambiguous  case ;  for  the  side  EG 
and  the  angle  at  E  are  respectively  of  the  same  affection 
with  the  angle  A  and  the  side  AG  (Art.  56) ;  and  the  hypo- 
thenuse  is  less  or  greater  than  a  quadrant,  according  as  AG 
and  GE  are  of  the  same  or  different  affections  (Art.  57). 


4.  Given,  one  leg  36°  45',  and  the  opposite  angle  42°  16', 
to  find  the  rest. 

Construction.  Describe 
the  primitive  circle  ABC, 
and  the  oblique  great  cir- 
cle  AC,  making   at   the 
point  A  an  angle  equal  to 
the  given  one  (Art.  162) ; 
i     about  the  pole  F  of  the 
primitive,  at   a  distance 
equal  to  the  complement 
of  the  given  leg,  describe 
a  less  circle,  cutting  the 
oblique   circle  AC  in  E 
(Art.   155);    through    E 
and  the  pole  F  describe  the  great  circle  GEF  (Art.  159), 
cutting  the  primitive  in  G ;  then  AGE  is  the  triangle  pro- 
posed. 

Calculation. 

As  sin  A  :  sin  EG  : :  rad  :  sin  AE  (Art.  58)  =    \    62°  49'' 

(  117°  11'., 


208 


SECTION   V. 


As  tan  A  :  tan  EG  : :  rad  :  sin  AG  (Art.  60)  =   j 
As  cos  EG  :  cos  A  : :  rad  :  sin  AEG  <Art.  67)  =   j 


55°  15'. 
124°  45'. 

67°  27'. 
112°  33'. 


The  cases  contained  in  this  problem  are  ambiguous,  as  is 
obvious  from  the  construction ;  the  unknown  sides  and  angle 
being  susceptible  of  two  values,  which  are  supplemental  to 
each  other. 


5.  Given,  the  two  legs,  70°  29'  and  30°  16',  to  find  the 
rest. 

Construction.  On  the 
primitive  circle  lay  down 
AG  equal  to  one  of  the 
legs  ;  and  through  G  and 
the  pole  of  the  primitive 
describe  the  right  circle 
GF ;  on  which  lay  down 
GE  equal  to  the  other  leg 
(Art.  169) ;  through  A  and 
E  describe  (by  Art.  159) 
the  great  circle  AEC; 
and  AGE  is  the  triangle 
which  was  to  be  con- 
structed. 

Calculation. 

As  sin  AG  :  rad  : :  tan  EG  :  tan  A  (Art.  60)  =  31°  46'. 

As  sin  EG  :  rad  : :  tan  AG  :  tan  E  (Art.  60)  =  79°  52'. 

As  rad  :  cos  AG  : :  cos  EG  :  cos  AE  (Art.  65)  ==  73°  14'. 

This  problem  contains  no  ambiguity ;  for  the  angles  at  A 
and  E  are  of  the  same  affections  as  the  opposite  sides  (Art. 
56)  ;  and  AE  is  less  than  a  quadrant,  when  AG  and  GE  are 
of  the  same  affection  (Art.  57). 


SPHERICAL  PROJECTIONS. 


209 


6.  Given,  the  two  oblique  angles,  28°  19'  and  75°  15',  to 
find  the  sides. 

Construction.  Describe  the  oblique  great  circle  AC,  making 

with  the  primitive  at  A 
an  angle  equal  to  one  of 
those  given  (Art.  162) ; 
about  the  pole  P  of  this 
circle,  at  a  distance  equal 
to  -  the  measure  of  the 
other  given  angle,  or  of 
its  supplement  if  obtuse, 
describe  a  less  circle  cut- 
ting the  primitive  in  H 
and  I  (Art.  157);  from  H 
nearest  to  A,  when  the 
second  angle  is  acute,  or 
from  I  the  more  remote 
point  of  intersection,when 
the  angle  is  obtuse,  lay  down  HG  on  the  primitive  equal  a 
quadrant;  through  G  describe  a  great  circle  at  right  angles 
to  AG,  cutting  AC  in  E  ;  then  AGE  is  the  triangle  proposed. 
For,  H  is  evidently  the  pole  of  GEF ;  and  the  distance  HP 
is  the  measure  of  the  angle  AEG. 

Calculation. 
As  tan  A  :  cotan  E  : :  rad  :  cos  AE  (Art.  64)  =  60°  45'. 
As  sin  A  :  rad  : :  cos  E  :  cos  AG  (Art.  67)      =  57°  32'. 
As  sin  E  :  rad  : :  cos  A  :  cos  EG  (Art.  67)      =  24°  27'. 

This  problem  contains  no  ambiguity;  for  the  sides  AG  and 
EG  are  of  the  same  affection  as  the  opposite  angles  (Art. 
56) ;  and  when  those  sides-,  or  their  opposite  angles,  are  of 
the  same  affection,  the  hypothenuse  is  less  than  a  quadrant 
(Art.  57). 


27 


u 


210  SECTION   V. 


Of  Oblique  Angled  Spherical  Triangles. 

1.  Given,  two  sides,  AC  45°  30',  BC  30°  30',  and  the  an^ 
A  opposite  one  of  them,  36°  45',  to  find  the  rest. 


Construction.  On  the  primitive  circle  lay  down  AC,  one 
of  the  given  sides ;  about  the  pole  C  describe  a  less  circle,  at 
the  distance  BC  of  the  other  given  side ;  describe  a  great 
circle,  making,  at  the  point  A,  with  the  primitive,  an  angle 
equal  to  the  given  one  (Art.  162),  intersecting  the  less  circle 
in  B ;  through  C  and  B  describe  a  great  circle  (Art.  159) ; 
and  ABC  is  the  triangle  to  be  made. 

When  BC,  the  side  opposite  the  given  angle,  is  the  less  of 
the  two,  there  may  be  two  points  of  intersection,  and,  conse- 
quently, two  positions  of  B.  Hence,  in  that  case,  the  problem 
is  ambiguous. 

Calculation:  Describe  the  great  circle  CD  through  the 
pole  of  AB;  then  ADC,  BDC  are  rectangular  triangles. 
Hence, 

As  rad  :  cos  A  : :  tan  AC  :  tan  AD  (Art.  62)  =       39°  12'. 

As  cos  AC  :  cos  BC  : :  cos  AD  :  cos  BD  (Art.  66)-  17°  41'. 


SPHERICAL   PROJECTIONS.  211 

As  rad  :  cos  AC  : :  tan  A  :  cotan  ACD  (Art.  64)=62°  22'. 

As  tan  BC  :  tan  AC  : :  cos  ACD  :  cos  BCD  (Art.  63)=36°  46'. 

As  sin  BC  :  sin  AC  : :  sin  A  :  sin  ABC  (Art.  59) =57°  14' 
or  122°  46'. 

Hence,  AB  =  56°  53',  or  21°  31' ;  and  ACB  =  99°  8',  or 
25°  36'. 

Or,  without  a  perpendicular. 
Find  ABC  as  above.     Then, 

Ascosi(AC  — BC)  :  cos  i(AC  +  BC)  ::  tan  i(ABC  +  BAC) 
:  cotan  JACB  (Art.  77,  eq.  15)=49°  34',  or  12°  48'. 

Hence,  ACB  =  99°  8',  or  25°  36'. 

And, 
As  cos  i(ABC  —  BAC)  :  cos  i(ABC  +  BAC)  ::  tan  -J(AC  + 

BC)  :  tan  £AB  (Art.  77,  eq.  17) =28°  26^',  or  10°  451'. 
Whence,  AB  56°  53',  or  21°  31'. 

2.  Given,  two  sides,  75°  20'  and  60°  16',  and  the  included 
angle  40°  18',  to  find  the  rest. 

Construction.  Describe 
a  great  circle,  making  at 
the  given  point  A  an 
angle  with  the  primitive 
equal  to  the  given  one 
(Art.  162).  Make  AB, 
AC,  respectively,  equal 
to  the  given  sides  (Art. 
169) ;  and  describe  a 
great  circle  through  B 
and  C  (Art.  159);  then 
ABC  is  the  triangle  pro- 
posed. 

In  this  problem,  there  is  no  ambiguity. 


212  SECTION    V. 

Calculation.  Draw  CD  at  right  angles  to  AB.     Then, 

As  rad  :  cos  A  ::  tan  AC  :  tan  AD  (Art.  62)  =  53°  10'. 
As  sin  BD  :  sin  AD  : :  tan  A  :  tan  B  (Art.  61)  =  60°  56'. 
As  cos  AD  :  cos  BD  : :  cos  AC  :  cos  BC  (Art.66)=39°  59'. 
As  rad  :  cos  AC  : :  tan  A  :  cotan  A  CD  (Art.  64)=  67°  11'. 
As  tan  AD  :  tan  BD  ::  tan  ACD  :  tan  BCD  (Art.  69)  =35°  57'. 
Hence,  ACB  =  103°  8'. 

Or,  without  a  perpendicular, 

As  cos  J(AB  +  AC)  :  cos  £(AB  —  AC)  ::  cot  £BAC 
:  tan  i(ACB  +  ABC)  (Art.  77,  eq.  11)  =  82°  2'. 

As  sin  i(AB  +  AC)  :  sin  |(AB  —  AC)  : :  cotan  £BAC 
:  tan  £(ACB— ■  ABC)  (Art.  77,  eq.  12)  ==  21°  6'. 

Whence,  ACB  =  103°  8',  and  ABC  =  60°  56'. 

As  sin  ABC  :  sin  BAC  : :  sin  AC  :  sin  BC  (Art.  59)=39°  59'. 

3.  Given,  one  side  80°  44',  and  the  two  adjacent  angles, 
40°  50'  and  70°  12',  to  find  the  rest. 

Construction.  On  the 
primitive  circle,  lay  down 
AB  equal  to  the  given 
side;  and  describe  two 
great  circles,  making  an- 
gles with  the  primitive, 
at  the  points  A  and  B, 
equal  to  the  given  ones 
(Art.  162),  and  let  those 
circles  cut  each  other  in 
C;  ABC  is  the  triangle 
proposed. 


SPHERICAL    PROJECTIONS.  213 

Calculation.  Through  B,  the  extremity  of  a  given  side, 
and  P  the  pole  of  AC,  describe  a  great  circle,  cutting  AC 
(produced,  if  necessary)  in  D.     Then, 

As  rad  :  cos  A  : :  tan  AB  :  tan  AD  (Art.  62)=  77°  50'. 
As  rad  :  cos  AB  : :  tan  A  :  cotan  ABD  (Art.  64)=  82°  5'. 
As  cos  CBD  :  cos  ABD  ::  tan  AB  :  tan  CB  (Art.  63)=40°  49'. 
As  tan  ABD  :  tan  CBD  ::  tan  AD  :  tan  CD  (Art.  69)=  7°  44'. 
As  sin  ABD  :  sin  CBD  ::  cos  A  :  cos  BCD  (Art.  68) =80°  57'. 
Hence,  AC  =  70°  6',  and  ACB  ==  99°  3'. 
Or,  without  a  perpendicular, 

As  cos  J  (B  +  A)  :  cos  i(B  —  A)  : :  tan  £AB  :  tan  £(AC  + 
BC)  (Art.  77,  eq.  13)  =  55°  27'. 

As  sin  i(B  +  A)  :  sin  J(B  —  A)  : :  tan  ^AB  :  tan  \(A&  — 
BC)  (eq.  14)  =  14°  39'. 

As  cos  !(AC  —  BC)  :  cos  i(AC  +  BC)  : :  tan  J(B  +  A) 

:  cot  JACB  (eq.  15)  =  49°  31'. 
Whence,  AC  =  70°  6',  BC  =  40°  48',  and  ACB  =  99°  2'. 

4.  Given,  two  angles,  50°  16'  and  60°  36',  and  a  side  42° 
34',  opposite  one  of  them,  to  find  the  rest. 

Construction.  Describe 
(Art.  162)  a  great  circle, 
making  with  the  primi- 
tive, at  the  point  A,  an 
angle  equal  to  the  given 
one,  to  which  the  given 
side  is  adjacent.  On  that 
circle  lay  down  AC  equal 
to  the  given  side  (Art. 
169)  ;  through  C  describe 
(by  Art.  163)  a  great 
circle,  making  with  the 
primitive  an  angle  equal 
28 


21-1 


SECTION   V. 


to  the  other  given  one ;  and  let  that  r.ircle  cut  the  primitive 
in  B  :  then  ABC  is  the  triangle. 

Calculation.  Draw  CD  at  right  angles  to  AB.  Then, 
As  rad  :  cos  A  : :  tan  AC  :  tan  AD  (Art.  62)=  30°  25'. 
As  tan  B  :  tan  A  : :  sin  AD  :  sin  BD  (Art.  61)=  20°  4'  21". 
As  rad  :  cos  AC  : :  tan  A  :  cotan  ACD  (Art.  64)=48°  27'  26". 
As  cos  A  :  cos  B  : :  sin  ACD  :  sin  BCD  (Art.  68) =35°  5'  9". 
As  sin  B  :  sin  A  : :  sin  AC  :  sin  BC  (Art.  59)=  36°  39'  46". 
Wherefore,  AB  =  50°  29'  21",  and  ACB  =  83°  32'  35". 

Or,  without  a  perpendicular. 
Fin»l  BC  as  above.     Then, 

As  cos  ^(B  —  A)  :  cos  J(B  +  A)  : :  tan  J(AC  +  BC) 
:  tan  £AB  (Art.  77,  eq.  17)  =  25°  14'  42". 

As  cos  i r(AC  —  BC)  :  cos  ^(AC  +  BC)  : :  tan  £(B  +  A)  : 
cotan  JACB  (eq.  15)  =  41°  46'  17". 

Wherefore,  AB  =  50°  29'  24",  and  ACB  =  83°  32'  34". 


5.  Given,  the  three  sides,  80c 
find  the  angles. 


16',  60°  44',  and  50°  20',  to 

Construction.  On  the 
primitive,  lay  down  AB 
equal  to  one  of  the  given 
sides ;  from  A  and  B,  as 
poles,  at  distances  equal 
to  the  other  sides  respec- 
tively, describe  (Art.  156) 
two  circles,  ICK  and 
GCH,  cutting  each  other 
in  C ;  through  A,  C,  and  B, 
C,  describe  (Art.  159)  two 
great  circles;  then  ABC 
is  the  triangle  proposed. 


SPHERICAL  PROJECTIONS.  ,  215 

Calculation.  Through  C,  describe  the  great  circle  CD  at 
right  angles  to  AB,  and  bisect  AB  in  E ,  then, 

As  tan  AE  :  tan  ^(AC  +  BC)  : :  tan  J(AC  —  BC)  :  tan  ED 

(Art.  74)  =  8°  56'  14". 
Whence,  AD  =  49°  4'  14",  and  BD  =  31°  11'  46". 
As  tan  AC  :  tan  AD  : :  rad  :  cos  A  (Art.  62)=49°  44'  18' . 
As  tan  BC  :  tan  BD  : :  rad  :  cos  B  =  59°  51'  33". 

As  sin  AC  :  sin  AD  : :  rad  :  sin  ACD  (Art.  58)  -  60°  0'  17". 
As  sin  BC  :  sin  BD  : :  rad  :  sin  BCD  =  42°  17'  25" 

Wherefore,  ACB  =  102°  17'  42". 

Or,  without  a  perpendicular. 

Take  P  =  half  the  sum  of  the  sides ;  then, 

As  sin  P.sin  (P— BC)  :  sin  (P  — AC).sin  (P  — AB)::  rad2 

:  tan2  £A  (Art.  77,  eq.  7)  «  24°  52'  9". 
Wherefore,  A  =  49°  44'  18". 
And  (Art.  59),  B  =  59°  51'  34",  and  C  =  102°  17'  42". 

6.  Given,  three  angles,  60°  36',  66°  20',  and  99°  50',  to  find 
the  sides. 

Construction.  At  the  centre  of  the  primitive,  make  an  angle 
C  equal  to  the  greatest  given  angle ;  and  describe  a  great 


216  SECTION  V. 

circle,  making,  with  the  right  circles  including  that  angle, 
two  angles  respectively  equal  to  the  other  two  given  ones 
(Art.  166) ;  and  let  A  and  B  be  those  angles :  then  ABC  is 
the  triangle  proposed. 

Calculation.  Through  C  let  two  great  circles,  CD  and  CE, 
pass ;  the  former  at  right  angles  to  AB,  and  the  latter  bisect- 
ing the  angle  ACB.     Then,  ^ 

As  cotan  £(B  +  A)  :  tan  KB"— A)  : :  tan  ACE  :  tan  ECD 
(Art.  75)  =  6°  47'  44". 

Whence,  ACD  =  56°  42'  44",  and  BCD  =  43°  7'  16". 

As  |an  A  :  cotan  ACD  : :  rad  :  cos  AC  ===  68°  17'  13". 

As  tan  B  :  cotan  BCD  : :  rad  :  cos  BC  (Art.  64) =62°    5'  42". 

As  sin  A  :  rad  : :  cos  ACD  :  cos  AD  (Art.  67)=    50°  57'    5". 

As  sin  B  :  rad  : :  cos  BCD  :  cos  BD=  37°    9'  41". 

AB  =  88°    6' 46". 
Or,  without  a  perpendicular, 

As  sin  B.sin  C  :  cos  |(A  +  C  —  B).cos  $(A  +  B  —  C)  : :  rad2 
:  cos2  ^BC  (Art.  77,  eq.  9) ; 

whence,  BC  =  62°  5'  58".  And  (Art.  59),  AB  =  8«8°  6'  46", 
AC  =  68°  17'  12". 

Otherwise.  From  A  and  B,  as  poles,  describe  the  great 
circles  FG,  FH,  cutting  the  primitive  in  G  and  H ;  then  the 
triangle  FGH  is  supplemental  to  ABC  (Art.  54).  That  is, 
GH  =  180°  —  ACB ;  FH  =  180°  —  ABC ;  FG  =  180°  — 
BAC  ;  whence  the  sides  of  FGH  are  known.     Then, 

As  sin  FH.sinFG  :  sin  i(HG  +  FH  —  FG).sin  i(HG  +  FG 
—  FH)  ::  rad2  :  sin2  JHFG  or  cos2  JAB ; 

whence,  AB  =  88°  6'  46",  as  before. 


SPHERICAL    PROJECTIONS. 


Promiscuous  Examples. 

The  following  astronomical  terms  being  occasionally  used 
in  the  succeeding  examples,  it  is  deemed  advisable  to  insert 
their  definitions. 

The  axis  of  the  earth  is  the  line  through  the  centre  on 
which  it  revolves;  the  points  where  the  axis  meets  the 
surface  of  the  earth  are  the  poles  of  the  earth ;  and  the  points 
where  the  axis  produced  meets  the  concave  surface  of  the 
visible  heavens,  are  the  celestial  poles. 

The  common  section  of  the  earth's  surface  and  a  plane 
passing  through  its  centre  at  right  angles  to  its  axis,  is 
termed  the  terrestrial  equator;  and  the  section  of  the  same 
plane  and  the  concave  surface  of  the  visible  heavens,  is  called 
the  celestial  equator,  or  equinoctial  circle. 

Meridians  are  great  circles  passing  through  the  poles,  and, 
consequently,  cutting  the  equator  at  right  angles. 

A  right  line  drawn  from  the  centre  of  the  earth,  through 
the  place  of  an  observer,  and  continued  till  it  meets  the  celes- 
tial sphere,  cuts  it  in  a  point  which  is  termed  the  zenith  of 
the  place.  The  point  where  the  same  line,  extended  beyond 
the  centre,  meets  the  celestial  sphere,  is  termed  the  nadir. 

The  great  circle  of  which  the  zenith  and  nadir  are  the 
poles,  is  termed  the  horizon. 

A  meridian  passing  through  the  zenith  of  any  place,  is 
called  the  meridian  of  that  place. 

The  angle  formed  by  the  meridian  of  a  place,  and  a  great 
circle  passing  through  the  zenith,  and  a  celestial  object,  is 
called  the  azimuth. 

The  distance,  reckoned  in  degrees,  minutes,  &c,  on  the 
meridian,  between  the  equator  and  a  place  on  the  earth's 
surface,  is  termed  the  latitude  of  that  place. 

The  angle  contained  between  the  meridian  of  a  place,  and 
28  ■     t 


213  SECTION    V 

some  other  assumed  as  a  first  meridian,  is  termed  the  longi- 
tude of  the  place.  The  longitude  is  usually  reckoned  east- 
ward or  westward  as  far  as  180°. 

The  common  section  of  the  plane  of  the  earth's  orbit  and 
the  celestial  sphere,  is  called  the  ecliptic.  This  circle  is  the 
sun's  apparent  annual  path. 

The  points  where  the  ecliptic  cuts  the  celestial  equator  are 
termed  the  equinoxes.  The  point  in  which  the  sun  appears 
when  passing  from  the  southern  to  the  northern  side  of  the 
equator,  is  termed  the  vernal  equinox. 

The  arc  of  the  celestial  equator,  reckoning  eastward, 
between  the  vernal  equinox  and  the  point  where  a  meridian 
through  any  celestial  object  cuts  the  equator,  is  called  the 
right  ascension  of  that  object ;  and  the  arc  of  the  meridian 
between  the  object  and  the  equator,  is  termed  the  decli- 
nation. 

The  arc  of  the  ecliptic,  reckoned  eastward,  between  the 
vernal  equinox  and  the  point  where  a  great  circle,  passing 
through  the  pole  of  the  ecliptic  and  any  celestial  object,  cuts 
the  ecliptic,  is  termed  the  longitude  of  that  object ;  and  the 
arc  of  that  great  circle,  betwreen  the  object  and  the  ecliptic, 
is  termed  its  latitude. 

The  angle  formed  by  the  ecliptic  and  the  celestial  equator 
is  termed  the  obliquity  of  the  ecliptic. 

The  right  ascension  of  the  point  in  the  celestial  equator 
which  is  cut  by  the  meridian  of  a  place,  is  called  the  right 
ascension  of  the  mid-heaven. 

The  point  of  the  ecliptic  which  is  cut  by  a  great  circle 
passing  through  its  pole  and  the  zenith  of  a  place,  is  called 
the  nonagesima  degree. 

Ex.  1.  Required,  the  distance  on  a  great  circle  of  the  earth 
between  Point  Venus  in  Otaheite,  and  Edinburgh ;  also,  the 
direction  of  each  from  the  other ;  the  latitude  of  the  former 
berns  17°  29'  South,  and  longitude  149°  29'  West;  and  the 


SPHERICAL   PROJECTIONS. 


219 


atitude  of  the  latter  55°  57'  North,  and  longitude  3°  11' 
West. 

Cons/ruction.  Assume  the 
primitive  circle  as  the  meri- 
dian of  Point  Venus ;  and 
take  P  as  the  north  pole; 
from  P  lay  down  PA  =  the 
polar  distance,  107°  29',  of 
the  place ;  through  P  describe 
a  great  circle,  making,  with 
the  primitive  at  P,  an  angle 
146°  18',  equal  to  the  differ- 
ence of  longitudes  (Art.  162) ;  on  this  circle  lay  down  PB 
34°  3'  =  the  polar  distance  of  Edinburgh  (Art.  169) ;  through 
A  and  B  describe  a  great  circle.  Then  AB  is  the  arc,  and 
PAB,  PBA  the  angles  required. 

Computation.  The  sides  AP,  BP,  and  the  angle  APB,  being 
given,  the  angles  at  A  and  B  are  found  (Art.  77,  eq.  11,  12), 
viz. :  PAB  =  25°  32',  PBA  =  47°  15'.  Then  (eq.  18),  AB  ** 
133°  53'.  Consequently,  Edinburgh  bears  from  Point  Venus, 
N.  25°  32'  E. ;  and  Point  Venus  bears  from  Edinburgh,  N. 
47°  15'  W. ;  distance  133°  53'. 


Ex.  2.  Required,  the  bearing  and  distance  on  a  great  circle 
of  the  Observatory  at  Greenwich  from  the  Capitol  at  Wash- 
ington: the  latitude  of  the  former  being  51°  28'  40"  N. ;  the 
latitude  of  the  latter  3S°  53'  N.,  and  longitude  77°  2'  W.  from 
the  meridian  of  Greenwich. 

Ans.  N.  49°  20'  E. ;  dist.  53°  8'. 

Ex.  3.  When  the  sun's  declination  is  23°  28'  N.,  how  long 
after  midnight  does  it  rise  in  latitude  39°  57'  N.,  and  how  far 
from  the  northern  point  of  the  horizon  ? 

The  construction  of  this   problem  is  readily  understood 


220 


SECTION   V. 


from  the  figure ;  taking  HO 
for  the  horizon;  OP=39°  57', 
the  latitude  of  the  place ;  de- 
scribing a  circle  about  the 
pole  P,  at  the  distance  66° 
32',  the  sun's  polar  distance ; 
supposing  this  circle  to  cut 
HO  in  n ;  then,  the  great  cir- 
cle PwS  being  described,  the 
triangle  POrc,  right  angled  at 
O,  will  contain  the  elements 
required ;  the  angle  ?vPO,  con- 
verted into  time  at  the  rate  of  one  hour  to  15°,  or  four  min- 
utes to  1°,  will  be  the  time  required ;  and  nO,  the  required 
distance  from  the  northern  part  of  the  horizon. 

Result:  On  58°  42',  0?n  68°  41'. 

Ex.  4.  Given,  the  latitude  of  the  place,  39°  56'  N. ;  decli- 
nation of  the  sun,  23°  28'  N. ;  and  zenith  distance,  60°  30' ; 
to  find  the  azimuth  and  polar  angle. 

To  construct  this  on  the 
plane  of  the  meridian,  take 
the  primitive  circle  HZPO 
for  the  meridian;  HO,  the 
horizon ;  Z,  the  zenith ;  P, 
the  north  pole ;  and,  there- 
fore, PO  =  the  latitude  of 
the  place.  About  P  describe 
a  circle  at  a  distance  =  66° 
32',  the  sun's  polar  distance 
(Art.  156) ;  and  about  Z,  at 
F  a   distance  =  60°  30',  the 

sun's  zenith  distance,  de- 
scribe another  circle,  cutting  the  former  in  N ;  then  describe 
great  circles  through  NZ  and  NP  (Art.  159) ;  and  in  the 
triangle  NPZ,  the  angle  PZN  is  the  azimuth,  and  ZPN  the 


SPHERICAL  PROJECTIONS. 


221 


polar  angle  required.     Those  angles  may  be  computed  by- 
Art.  77,  eq.  5,  6,  or  7. 

Result:  NZP  82°  56',  ZPN  70°  20'. 

Ex.  5.  In  the  beginning  of  1842,  the  right  ascension  of 
Aldebaran  (the  bull's  eye)  was  66°  42'  50",  and  the  declina- 
tion 16°  8'  36"  N.  Required,  the  longitude  and  latitude  at 
that  time ;  the  obliquity  of  the  ecliptic  being  23°  27'  40". 

To  project  this  on  the 
plane  of  the  celestial  equa- 
tor, take  the  primitive 
ABCD  to  denote  that  cir- 
cle; A  and  C  being  the 
equinoxes,  and  P  the  north 
pole  :  through  A,  C,  describe 
the  circle  AGCH,  making 
an  angle  of  23°  27'  40"  with 
the  primitive ;  then  AGCH 
will  represent  the  ecliptic ; 
take  Q  the  pole  of  AGCH ; 
make  AE  =  the  star's  right 
ascension;  through  E,  P, 
draw  the  line  EP ;  this  will  represent  the  meridian  passing 
through  the  star.  On  EP  lay  down  ES  (Art.  169)  =  the 
star's  declination ;  then  S  is  the  place  of  the  star.  Through 
SQ  describe  (Art.  159)  a  great  circle  cutting  the  circle  AGC 
in  F ;  then  AF  is  the  longitude,  and  FS  the  latitude  required. 
Through  AS  describe  a  great  circle;  then  the  angle  EAS 
may  be  computed  by  Art.  60,  and  thence  AF  and  FS  by 
Arts.  71  and  70. 

Result :  long.  67°  34'  23" ;  lat.  5°  31'  18"  S. 

Ex.  6.  Given,  the  latitude  of  the  place,  40°  north ;  the 
obliquity  of  the  ecliptic,  23°  28' ;  and  the  right  ascension  of 
the  mid-heaven,  60° ;  to  find  the  longitude  and  altitude  of  the 
nonagesima  degree. 

29 


222 


SECTION    V. 


To  project  this  example  on  the  plane  of  the  meridian,  let 
the  primitive  circle  HZMO  denote  the  meridian;  HO,  the 
horizon;  Z,  the  zenith,  or  pole  of  HO.  On  the  primitive 
circle  lay  down  HP,  ZM,  each  equal  to  the  latitude  of  the 
place,  or  40°  ;  then  P  will  be  the  pole  of  the  equator.  About 
P,  at  the  distance  23°  28',  the  obliquity  of  the  ecliptic,  de- 
scribe a  less  circle;  through  M  and  C  (the  centre  of  the 
primitive),  draw  the  right  circle  CMA ;  this  will  represent 
the  equator.  Make  MA  =  60°,  the  right  ascension  of  the 
mid-heaven;  the  point  A  will  be  the  vernal  equinox.  Make 
CL  =  30° ;  then  L  will  be  the  autumnal  equinox,  for  CM  = 
90°.  About  the  pole  L  describe  the  great  circle  QPS,  cut- 
ting the  less  circle  in  Q ;  then  Q  is  the  pole  of  the  ecliptic. 
Through  L  describe  a  great  circle  having  Q  for  its  pole ;  this 
circle  will  denote  the  ecliptic,  and  pass  through  A.  Lastly, 
through  Q  and  Z  describe  a  great  circle  QZNR,  cutting  the 


SPHERICAL    PROJECTIONS. 


223 


ecliptic  in  N,  and  the  horizon  in  R ;  N  is  the  nonagesima, 
whose  longitude  is  AN,  and  altitude  NR  or  QZ. 

Calculation,  Let  QP  cut  the  equator  in  E  ;  then  AE — AM 
=EM=ZPE»  Hence,  QPZ  becomes  known  =  150°;  then, 
in  the  triangle  ZPQ,  ZP  ==  50°,  PQ  =  23°  28' ;  with  which 
and  the  contained  angle,  we  may  find  the  angle  ZQP  =  23° 
53'  44",  and  side  QZ  ==  NR  =  71°  0'  18".  Consequently,  the 
longitude  of  the  point  N  =  66°  6'  16". 

Ex.  7.  Given,  the  latitude  of  the  place  41°  36'  north,  anc4 
the  sun's  declination  22°  10'  north,  to  find  the  time  from  noon, 
or  the  polar  angle,  when  the  sun  is  on  the  vertical  circle 
which  passes  through  the  east  and  west  points  of  the  horizon, 
and  the  sun's  altitude  at  the  same  time. 

Result:  Polar  angle  62°  41',  or  time  from  noon  4  hours  11 
minutes  ;  altitude  34°  38'. 

Ex.  8.  On  the  first  day  of  the  year  1836,  when  it  was  noon 
at  Greenwich,  the  right  ascension  of  Jupiter  was  101°  55', 
and  declination  23°  4'  2"  north ;  at  the  same  time,  the  right 
ascension  of  Saturn  was  212°  17',  and  the  declination  10°  30' 
50"  south.  What  was  their  distance  on  the  arc  of  a  great 
circle  1 

Ans.  112°  44'  59'. 


Ex.  9.  In  the  beginning 
of  1842,  the  declination  of 
the  polar  star  was  88°  28' 
north.  What  was  its  azi- 
muth, when  its  elongation 
from  the  meridian  was  the 
greatest,  the  observer  being 
in  latitude  40°  north  ? 

Ans.  2°  0'  6". 


224 


SECTION   V. 


Ex.  10.  In  latitude  40°  north,  required  the  duration  of 
twilight  at  the  several  times  when  the  sun's  declination  is 
23°  28'  south,  5°  50'  south,  and  23°  28'  north ;  the  twilight 
being  supposed  to  begin  when  the  sun's  centre  is  49'  below 
the  horizon,*  and  to  end  when  it  is  18°  below. 

Ans.  1  h.  35  m. ;  1  h.  29  m.  ;f  and  2  h.  4  m. 

Ex.  11.  Given,  two  zenith  distances  of  the  sun's  centre, 
65°  20'  and  60°  18',  taken  at  the  same  place,  both  being  in 
the  forenoon ;  the  interval  between  the  observations,  mea- 
sured by  a  good  time-piece,  1  hour  32  minutes ;  the  sun's 
declination  20°  south,  and  the  approximate  latitude  of  the 
place  40°  15'  north ;  to  find  the  time  of  the  last  observation, 
and  the  correct  latitude  of  the  place. 


Let  HO  be  the  horizon; 
HZO,  the  meridian ;  Z,  the 
zenith;  P,  the  north  pole; 
BZ  and  AZ,  the  given  zenith 
distances.  Then  PZ  is  the 
colatitude ;  and  PB,  PA,  the 
sun's  polar  distance. 


Now  (by  Art.  77,  eq.  1), 

cos  AZ  =  cos  ZPA.  sin  AP 
sin  ZP  +  cos  AP.cos  ZP ; 


and 


cos  BZ  =  cos  ZPB.sin  BP.sin  ZP  +  cos  BP.cos  ZP. 
Hence  (AP  being  =  BP), 
cos  AZ  —  cos  BZ  =  (cos  ZPA  —  cos  ZPB)  sin  AP.sin  ZP ; 


*  The  refraction  of  the  sun's  light,  when  in  the  horizon,  is  33',  and 
apparent  semi-diameter  about  16';  hence  his  upper  limb  is  just  visible 
when  the  centre  is  49'  below  the  horizon. 

j-  This  is  the  shortest  twilight  in  latitude  40°  N.,and  occurs  twice  in  the 
year,  viz.,  in  spring  and  »itumn.  when  the  sun's  declination  is  5°  50'  S. 


SPHERICAL   PROJECTIONS.  225 

consequently  (Art.  37,  eq.  13), 

sin  J(BZ  +  AZ).sin  J(BZ  —  AZ)  =  sin  J(BPZ  +  APZ). 
sin  i(BPZ  —  APZ)  sin  AP.sin  ZP ; 

of  which,  BPZ — APZ  is  given  from  the  elapsed  time.    Hence, 

•    imp7  m  AP7^      Bin«BZ  +  AZ).sini(BZ- AZ) 
sin  i(BPZ  +  APZ)  =  ^jxjfcye^  ^BPZ^-A JZ)  = 

15°  51'  11". 

Whence  APZ  -  4°  21'  11",  and  (by  Art.  59)  AZP  ==  175° 
17'  24"  ;  from  which  we  find  (by  Art.  77,  eq.  18),  ZP  ==-  49° 
50' ;  and  therefore  PO,  the  corrected  latitude,  40°  10'  ;»  and 
time  from  noon,  when  the  least  zenith  distance  was  taken,  17 
minutes  24  seconds. 

Ex.12.  Given,  the  approximate  latitude,  39°  26'  N.;  the 
sun's  declination,  20°  41'  N. ;  sun's  corrected  zenith  distance 
at  11  h.  30'  15",  by  watch,  21°  30' ;  and  at  12  h.  26'  28",  by 
watch,  18°  52'.  Required,  the  corrected  latitude,  and  error 
of  the  watch. 

Ans.  Lat.  39°  29' ;  watch  too  fast,  18  min.  57  sec. 

Ex.  13.  At  the  time  when  Sirius  and  Aldebaran  were  in 
the  same  vertical  circle,  the  true  zenith  distance  of  the  latter 
was  found  to  be  30°  16' ;  the  stars  being  on  the  east  of  the 
meridian.  Required,  the  latitude  of  the  place,  and  right 
ascension  of  the  mid-heaven ;  the  right  ascension  of  Sirius 
being  99°  33'  30",  and  its  declination  16°  30'  18"  south ;  the 
right  ascension  of  Aldebaran  66°  43'  45",  and  its  declination 
16°  11'  12"  north. 

Ans.  Lat.  35°  8'  42"  N. ;  right  ascension  of  mid-heaven 
39°  17'  23". 


*  When  the  latitude  thus  found  differs  considerably  from  the  approxi- 
mate latitude,  the  computation  ought  to  be  repeated,  with  the  result  first 
obtained  substituted  for  the  approximate  latitude. 

29 


226 


SECTION   V. 


Examples  of  a  Mixed  Character. 

Ex.  1.  From  the  top  of  a  cliff  near  a  river,  two  buoys  at 
anchor  being  observed,  whose  distance  from  each  other  was 
known  to  be  300  yards,  their  angles  of  depression  below  the 
plane  of  the  observer  were  found  to  be  30°  and  40°  respec- 
tively; and  the  angle  at  the  eye,  subtended  by  the  line  joining 
them,  was  37°.  Required,  the  distance  of  each  buoy  from 
the  observer,  and  the  altitude  of  the  cliff  above  the  level  of 
the  water. 

The  observed  depressions,  each  increased  by  90°,  form  two 
sides ;  and  the  angular  distance,  the  base  of  a  spherical  tri- 
angle, with  which  the  angle  opposite  the  base  is  found  =44°. 
This  is  the  horizontal  angle,  subtended  by  the  line  joining  the 
buoys. 

Drawing  then  a  vertical  line  through  the  position  of  the 
observer  to  meet  the  plane  of  the  water ;  and,  from  the  point 
where  it  meets  that  plane,  drawing  lines  to  the  buoys ;  those 
lines  will  be  to  each  other  as  the  cotangents  of  the  given 
angles  of  depression ;  and  the  angle  which  they  make  with 
each  other  will  be  44°,  as  above  found.  The  construction  is 
this : 


Take  AB=300,  the  given  dis- 
tance ;  on  AB  describe  a  segment 
of  a  circle  ACB,  containing  an 
angle  of  44° ;  complete  the  cir- 
cle, and  bisect  the  arc  AEB  in 
E;  make  the  angles  ABF  and 
BAF  =  30°  and  40°  respective- 
ly ;  draw  FG  at  right  angles  to 
AB;  join  EG,  and  produce  it  to 
meet  the  circle  in  C;  join  CA, 
CB ;  then,  since  the  angle  ACB  is  bisected  by  the  line  CG, 

As  AC  :  CB  : :  AG  :  BG  (3.6)  : :  cot  BAF  :  cot  ABF. 


SPHERICAL   PROJECTIONS.  227 

Consequently,  C  is  the  point  in  the  plane  of  the  water  which 
is  cut  by  a  vertical  line  passing  through  the  place  of  the 
observer. 

The  calculation  is  easily  made.     For  (Art.  28  and  Art.  37, 
eq.  8), 
As  sin  (BAF+ ABF)  :  sin  (BAF  —  ABF)  : :  AB  :  BG— AG. 

Hence  AG  is  known.  If  we  join  AE,  BE,  the  side  AB,  and 
all  the  angles  of  the  triangle  ABE,  are  given ;  whence  AE 
becomes  known.  Then,  in  the  triangle  AGE,  the  sides  AE, 
AG,  and  the  contained  angle,  are  known;  from  which  the 
angle  AEG  =  ABC  is  found.  In  the  triangle  ABC,  we  then 
have  the  base  AB  and  all  the  angles,  to  find  AC  and  BC. 
Then,  from  either  of  these  and  the  angle  of  depression,  the 
altitude  of  the  cliff  may  be  found.  Lastly,  with  the  distances 
AC,  BC,  and  the  angles  of  depression,  the  distances  from  A 
and  B  to  the  place  of  the  observer  are  determined 

Result :  Altitude  of  cliff,  249 ;  distances,  388  and  49S 

Ex.  2.  Suppose  an  observer  on  a  frozen  lake  takes  the 
altitudes  and  angular  distance  of  two  cliffs  on  the  shore,  as 
follows :  altitude  of  first,  50° ;  of  second,  55°  30' ;  angular 
distance,  25°  20'.  Then,  advancing  on  the  ice  500  yards,  in 
the  vertical  plane  which  passes  through  the  first  cliff,  the 
altitudes  are  57°  and  59°  respectively.  Required,  the  dis- 
tance of  the  cliffs  from  <Mkcfi  utfier,  and  their  respective 
altitudes  above  the  surface  of  the  lake. 

Ans.  Distance  of  cliffs,  2347  or  2850  yards ;  altitude  of 
first,  2636  yards ;  of  second,  4069  or  552  yards. 

Ex.  3.  The  crew  of  a  vessel  at  sea  discovering  a  light  in 
the  horizon,  which  they  suppose  to  be  a  vessel  on  fire,  sail 
directly  towards  it,  over  1°  of  a  great  circle,  when  they  per- 
ceive that  it  is  a  fire  on  a  mountain,  which  is  then  1°  30' 
above  the  horizon.  Required,  the  distance  of  the  light  when 
first  seen,  and  the  height  of  the  mountain  above  the  level  of 


228  SECTION   V. 

the  sea ;  the  earth  being  considered  as  a  spnere  wnose  radius 
is  3968  miles. 

Answer.  Distance,  138.5  miles ;  height,  2.4  miles. 

Ex.  4.  Given,  AB,  a  horizontal  line,  1785  yards  in  length, 
running  exactly  north ;  D,  C,  two  elevated  peaks,  eastward 
from  AB,  such  that  the  elevation  of  C  above  the  plane  of  the 
horizon,  seen  from  A,  is  16°  30',  and  the  elevation  of  D  20° 
40'.  But,  seen  from  B,  the  elevation  of  C  is  14°  25',  and  the 
elevation  of  D  13°  15'.  Also,  the  angle  BAG,  taken  in  the 
oblique  plane  passing  through  AB  and  C,  is  38°  16' ;  the 
angle  CAD,  taken  in  the  plane  which  passes  through  A,  C 
and  D,  87°  20'.  Required,  the  distance  and  bearing  of  DC 
when  reduced  to  the  horizontal  plane  on  which  AB  lies. 

Result :  DC,  N.  29°  W. ;  2,556  yards. 


TUB    END 


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JUN    12  1944 


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